As for mathematics, you should memorize and understand the following formula: the cross double method has no formula, if you must say it.
Is to use x 2+(p+q) x+PQ = (x+q) (x+p) where PQ is a constant. X 2 is the square of X.
1. Factorization
That is, the sum-difference product, the final result should be decomposed to no more points. Moreover, it is certain that if a polynomial can be decomposed into factors, the result will be unique, because if the polynomial f(x) with a degree greater than zero in the number field f is excluded, then f(x) can be uniquely decomposed into the following form:
F(x) = AP1k1(x) p2k2 (x) … piki (x) *, where α is the coefficient of the highest term of f (x), P 1 (x), P2 (x) … PI (x) is
(*) or the typical decomposition of polynomial f(x). Proof: See Gao Dai P52-53.
In elementary mathematics, the decomposition of polynomials is called factorization, and its general steps are: one mention, two sets, three groups and so on.
The requirement is: until we can't separate.
2. Method introduction
2. 1 common factor method:
If every term of a polynomial has a common factor, we can first consider putting forward the common factor and factorizing it. Note that every term must have a common factor.
Example 15x3+ 10x2+5x
Obviously, each term contains a common factor 5x, so we can consider extracting the common factor 5x, and the remaining x2+2x+ 1 can still be decomposed.
Solution: Original formula =5x(x2+2x+ 1)
=5x(x+ 1)2
2.2 formula method
That is, if a polynomial satisfies the structural characteristics of a special formula, it can be factorized by a set of formulas, so it is required to be familiar with some commonly used formulas. In addition to the basic formulas in textbooks, some basic formulas that often appear in math competitions are summarized as follows:
a2-b2=(a+b)(a-b)
a2 2ab+b2=(a b)2
a3+b3=(a+b)(a2-ab+b2)
a3-b3=(a-b)(a2+ab+b2)
a3 3a2b+3ab2 b2=(a b)3
a2+b2+c2+2ab+2bc+2ac=(a+b+c)2
a 12+a22+…+an2+2a 1 a2+…+2an- 1an =(a 1+a2+…+an)2
a3+B3+C3-3 ABC =(a+b+c)(a2+B2+C2-a B- AC-BC)
An+bn = (a+b) (an-1-an-2b+…+bn-1) (n is an odd number)
Explain the factorial theorem, that is, for unary polynomial f(x), if f(b)=0, it must contain linear factorial X-B. It can be judged that when n is even, when a = b and a =-b, there is an-bn=0, so an-bn must contain a+b and a-b factors.
Example 2 factorization: ① 64x6-y12②1+x+x2+…+x15.
Formulas can be used to analyze various minor problems.
Solution ① 64x6-y12 = (8x3-y6) (8x3+y6)
=(2x-y2)(4x 2+2x 2+y4)(2x+y2)(4x 2-2x 2+y4)
② 1+x+x2+…+x 15=
=( 1+x)( 1+x2)( 1+x4)( 1+x8)
When we pay attention to polynomial decomposition, we should first construct the formula and then decompose it.
2.3 Grouping decomposition method
When there are many terms in polynomials, polynomials can be grouped reasonably to achieve the purpose of smooth decomposition. Of course, other sub-methods can also be integrated, and the grouping method is not necessarily unique.
Example 1 decomposition factor: x15+m12+M9+M6+m3+1.
The formula = (x15+m12)+(M9+M6)+(m3+1).
= m 12(m3+ 1)+M6(m3+ 1)+(m3+ 1)
=(m3+ 1)(m 12+M6 ++ 1)
=(m3+ 1)[(M6+ 1)2-M6]
=(m+ 1)(m2-m+ 1)(M6+ 1+m3)(M6+ 1-m3)
Example 2 Factorization: x4+5x3+ 15x-9
Analysis can be grouped according to coefficient characteristics.
The solution formula =(x4-9)+5x3+ 15x.
=(x2+3)(x2-3)+5x(x2+3)
=(x2+3)(x2+5x-3)
2.4 cross multiplication
Cross multiplication can be considered for quadratic trinomial with the structural characteristics of ax2+bx+c,
That is, x2+(b+c)x+bc=(x+b)(x+c) When the coefficient of the x2 term is not 1, cross multiplication can also be performed.
Example 3 decomposition factor: ①x2-x-6②6x2-x- 12.
Solution ① 1x2
1x-3
Original formula =(x+2)(x-3)
②2x-3
3x4
Original formula =(2x-3)(3x+4)
Note: "ax4+bx2+c" type can also consider this method.
2.5 Pairwise cross multiplication
Cross multiplication is a basic method commonly used in quadratic trinomial decomposition. For more complex polynomials, especially some quadratic sextuples, such as 4x2-4xy-3y2-4x+ 10y-3, cross multiplication factorization can also be used. The specific steps are as follows:
(1) The quadratic trinomial composed of the first three times is decomposed by cross multiplication, and a cross multiplication graph is obtained.
(2) Divide the constant term into two factors and fill them in the right side of the second cross, so that the sum of the products of the intersection of the two factors in the second cross is equal to the primary term containing Y in the original formula, and the sum of the products of the intersection of the two factors at the left side of the first cross must be equal to the primary term containing X in the original formula.
Example 5 Factorization
①4x 2-4xy-3 y2-4x+ 10y-3②x2-3xy- 10y 2+x+9y-2
③a b+ B2+a-b-2④6x 2-7xy-3 y2-xz+7yz-2z 2
Solution ① Original formula =(2x-3y+ 1)(2x+y-3)
2x-3y 1
2xy-3
② Original formula =(x-5y+2)(x+2y- 1)
x-5y2
x2y- 1
③ Original formula =(b+ 1)(a+b-2)
0ab 1
ab-2
④ Original formula =(2x-3y+z)(3x+y-2z)
2x-3yz
3x-y-2z
Note: Type ③ can be supplemented by oa2, and double cross multiplication can be used. Of course, this question can also be grouped.
Such as (ab+a)+(B2-b-2) = a (b+1)+(b+1) (b-2) = (b+1) (a+b-2).
(4) The three letters of the formula satisfy the quadratic six-term formula, and -2z2 can be regarded as a constant decomposition:
2.6 Demolition method and addition method
For some polynomials, if factorization cannot be performed directly, one of them can be decomposed into the difference or sum of two terms. Then the factors are decomposed by grouping method and formula method, among which the method of splitting and adding items is not unique, and there are many different ways to solve them. Be sure to analyze the problem in detail and choose a simple decomposition method.
Example 6 Decomposition factor: x3+3x2-4
Analysis method 1: -4 can be decomposed into-1, and -3 is (x3- 1)+(3x2-3).
Method 2: Add x4 and subtract x4, that is, (x4+3x2-4)+(x3-x4).
Method 3: Add 4x and subtract 4x, that is, (x3+3x2-4x)+(4x-4).
Method 4: Divide 3x2 into 4x2-x2, namely (x3-x2)+(4x2-4).
Method 5: Break x3 into 4x2-3x3 (4x3-4)-(3x3-3x2) and so on.
Solution (Option 4) Original formula =x3-x2+4x2-4
= x2(x- 1)+4(x- 1)(x+ 1)
=(x- 1)(x2+4x+4)
=(x- 1)(x+2)2
2.7 alternative methods
The substitution method introduces a new letter variable, replaces the letter variable in the original formula, and simplifies the formula. Use this
This method can simplify the factorization of some special polynomials.
Example 7 factorization:
(x+ 1)(x+2)(x+3)(x+4)- 120
It will be very boring to analyze this, but we notice that
(x+ 1)(x+4)=x2+5x+4
(x+2)(x+3)=x2+5x+6
Therefore, the problem can be decomposed by substitution method.
Solution formula =(x2+5x+4)(x2+5x+6)- 120.
Let y=x2+5x+5, then the original formula = (y-1) (y+1)-120.
=y2- 12 1
=(y+ 1 1)(y- 1 1)
=(x2+5x+ 16)(x2+5x-6)
=(x+6)(x- 1)(x2+5x+ 16)
Note: You can also use x2+5x+4=y or x2+5x+6=y or x2+5x=y here. Please compare carefully which method is simpler?
2.8 undetermined coefficient method
The undetermined coefficient method is an important method to solve the deformation of algebraic constants. If the letter frame after algebraic deformation can be determined, but the letter coefficient is too high to be determined, we can first express the letter coefficient with unknowns, and then list n equations (groups) with special determined coefficients according to the properties of polynomial constants, and solve the equations (groups) to get the undetermined coefficients. The undetermined coefficient method is widely used, and only some applications of its factorization are studied here.
Example 7 Factorization: 2a2+3ab-9b2+ 14a+3b+20.
The analysis belongs to the quadratic six-term formula, and double cross multiplication can also be considered. Here we use the undetermined coefficient method.
Firstly, decompose 2a2+3ab+9b2=(2a-3b)(a+3b).
The solution can be set to the original formula =(2a-3b+m)(a+3b+n).
= 2 a2+3 ab-9 B2+(m+2n)a+(3m-3n)b+ Mn…………
Compare the coefficients of two polynomials (original formula and * formula)
m+2n= 14( 1)m=4
3m-3n=-3(2)= >
mn=20(3)n=5
∴ Original formula =(2x-3b+4)(a+3b+5)
Note that for formula (*), because the equations in which a and b take arbitrary values are valid, it is also possible to find m and n by the special value method.
Let a= 1, b=0, m+2n= 14m=4.
= & gt
Let a=0, b= 1, m=n=- 1n=5.
2.9 factorial theorem, comprehensive division factorization
For integral coefficient unary polynomial f (x) = anxn+an-1xn-1+…+a1x+A0.
According to the factorial theorem, it can be judged whether there is a first-order factorial (x-) (where P and Q are coprime), where P is the divisor of the first coefficient an and Q is the divisor of the last coefficient a0.
If f()=0, there must be (x-) to divide the polynomial synthetically.
Example 8 Factorization Factor x3-4x2+6x-4
This is an integer coefficient unary polynomial, because the positive divisor of 4 is 1, 2,4.
The possible factors are x1,x 2, x 4,
∫f( 1)≠0，f( 1)≠0
But f(2)=0, so (x-2) is the factor of this polynomial, and then divide by synthesis.
2 1-46-4
2-44
1-220
So the original formula =(x-2)(x2-2x+2).
Of course, this problem can also be decomposed, such as x3-4x2+4x+2x-4.
=x(x-2)2+(x-2)
=(x-2)(x2-2x+2)
There are many ways to decompose factors, and their methods are interrelated. A problem is likely to be solved by multiple methods at the same time. So after knowing these methods, we must pay attention to the flexible use of various methods and firmly grasp them!
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I don't know what textbook you are.
I will give you everything from junior high school.
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1 There is only one straight line at two points.
The line segment between two points is the shortest.
The complementary angles of the same angle or equal angle are equal.
The complementary angles of the same angle or the same angle are equal.
One and only one straight line is perpendicular to the known straight line.
Of all the line segments connecting a point outside the straight line with points on the straight line, the vertical line segment is the shortest.
7 Parallel axiom passes through a point outside a straight line, and there is only one straight line parallel to this straight line.
If both lines are parallel to the third line, the two lines are also parallel to each other.
The same angle is equal and two straight lines are parallel.
The dislocation angles in 10 are equal, and the two straight lines are parallel.
1 1 are complementary and two straight lines are parallel.
12 Two straight lines are parallel and have the same angle.
13 two straight lines are parallel, and the internal dislocation angles are equal.
14 Two straight lines are parallel and complementary.
Theorem 15 The sum of two sides of a triangle is greater than the third side.
16 infers that the difference between two sides of a triangle is smaller than the third side.
The sum of the internal angles of 17 triangle is equal to 180.
18 infers that the two acute angles of 1 right triangle are complementary.
19 Inference 2 An outer angle of a triangle is equal to the sum of two non-adjacent inner angles.
Inference 3 The outer angle of a triangle is greater than any inner angle that is not adjacent to it.
2 1 congruent triangles has equal sides and angles.
Axiom of Angular (SAS) has two triangles with equal angles.
The Axiom of 23 Angles (ASA) has the congruence of two triangles, which have two angles and their sides correspond to each other.
The inference (AAS) has two angles, and the opposite side of one angle corresponds to the congruence of two triangles.
The axiom of 25 sides (SSS) has two triangles with equal sides.
Axiom of hypotenuse and right angle (HL) Two right angle triangles with hypotenuse and right angle are congruent.
Theorem 1 The distance between a point on the bisector of an angle and both sides of the angle is equal.
Theorem 2 is a point with equal distance on both sides of an angle, which is on the bisector of this angle.
The bisector of an angle 29 is the set of all points with equal distance to both sides of the angle.
The nature theorem of isosceles triangle 30 The two base angles of isosceles triangle are equal (that is, equilateral and equiangular).
3 1 Inference 1 The bisector of the vertices of an isosceles triangle bisects the base and is perpendicular to the base.
The bisector of the top angle, the median line on the bottom edge and the height on the bottom edge of the isosceles triangle coincide with each other.
Inference 3 All angles of an equilateral triangle are equal, and each angle is equal to 60.
34 Judgment Theorem of an isosceles triangle If a triangle has two equal angles, then the opposite sides of the two angles are also equal (equal angles and equal sides).
Inference 1 A triangle with three equal angles is an equilateral triangle.
Inference 2 An isosceles triangle with an angle equal to 60 is an equilateral triangle.
In a right triangle, if an acute angle is equal to 30, the right side it faces is equal to half of the hypotenuse.
The center line of the hypotenuse of a right triangle is equal to half of the hypotenuse.
Theorem 39 Is the distance between the point on the vertical line of a line segment and the two endpoints of this line segment equal?
The inverse theorem and the point where the two endpoints of a line segment are equidistant are on the middle vertical line of this line segment.
The perpendicular bisector of the 4 1 line segment can be regarded as the set of all points with equal distance from both ends of the line segment.
Theorem 42 1 Two graphs symmetric about a line are conformal.
Theorem 2: If two figures are symmetrical about a straight line, then the symmetry axis is the perpendicular to the straight line connecting the corresponding points.
Theorem 3 Two graphs are symmetrical about a straight line. If their corresponding line segments or extension lines intersect, then the intersection point is on the axis of symmetry.
45 Inverse Theorem If the straight line connecting the corresponding points of two graphs is bisected vertically by the same straight line, then the two graphs are symmetrical about this straight line.
46 Pythagorean Theorem The sum of squares of two right angles A and B of a right triangle is equal to the square of the hypotenuse C, that is, A 2+B 2 = C 2.
47 Inverse Theorem of Pythagorean Theorem If the three sides of a triangle A, B and C are related in length A 2+B 2 = C 2, then the triangle is a right triangle.
The sum of the quadrilateral internal angles of Theorem 48 is equal to 360.
The sum of the external angles of the quadrilateral is equal to 360.
The theorem of the sum of internal angles of 50 polygons is that the sum of internal angles of n polygons is equal to (n-2) × 180.
5 1 It is inferred that the sum of the external angles of any polygon is equal to 360.
52 parallelogram property theorem 1 parallelogram diagonal equality
53 parallelogram property theorem 2 The opposite sides of parallelogram are equal
It is inferred that the parallel segments sandwiched between two parallel lines are equal.
55 parallelogram property theorem 3 diagonal bisection of parallelogram.
56 parallelogram decision theorem 1 Two groups of parallelograms with equal diagonals are parallelograms.
57 parallelogram decision theorem 2 Two groups of parallelograms with equal opposite sides are parallelograms.
58 parallelogram decision theorem 3 A quadrilateral whose diagonal is bisected is a parallelogram.
59 parallelogram decision theorem 4 A group of parallelograms with equal opposite sides are parallelograms.
60 Rectangle Property Theorem 1 All four corners of a rectangle are right angles.
6 1 rectangle property theorem 2 The diagonals of rectangles are equal
62 Rectangular Decision Theorem 1 A quadrilateral with three right angles is a rectangle.
63 Rectangular Decision Theorem 2 Parallelograms with equal diagonals are rectangles
64 diamond property theorem 1 all four sides of the diamond are equal.
65 Diamond Property Theorem 2 Diagonal lines of diamonds are perpendicular to each other, and each diagonal line bisects a set of diagonal lines.
66 Diamond area = half of diagonal product, that is, S=(a×b)÷2.
67 diamond decision theorem 1 A quadrilateral with four equilateral sides is a diamond.
68 Diamond Decision Theorem 2 Parallelograms whose diagonals are perpendicular to each other are diamonds.
69 Theorem of Square Properties 1 All four corners of a square are right angles and all four sides are equal.
Theorem of 70 Square Properties 2 Two diagonal lines of a square are equal and bisected vertically, and each diagonal line bisects a set of diagonal lines.
Theorem 7 1 1 is congruent with respect to two centrosymmetric graphs.
Theorem 2 About two graphs with central symmetry, the connecting lines of symmetric points both pass through the symmetric center and are equally divided by the symmetric center.
Inverse Theorem If a straight line connecting the corresponding points of two graphs passes through a point and is bisected by the point, then the two graphs are symmetrical about the point.
The property theorem of isosceles trapezoid is that two angles of isosceles trapezoid on the same base are equal.
The two diagonals of an isosceles trapezoid are equal.
76 isosceles trapezoid decision theorem A trapezoid with two equal angles on the same base is an isosceles trapezoid.
A trapezoid with equal diagonal lines is an isosceles trapezoid.
Theorem of bisecting line segments by parallel lines If a group of parallel lines are tangent to a straight line.
Equal, then the line segments cut on other straight lines are also equal.
79 Inference 1 A straight line passing through the midpoint of one waist of a trapezoid and parallel to the bottom will bisect the other waist.
Inference 2 A straight line passing through the midpoint of one side of a triangle and parallel to the other side will bisect the third side.
The median line theorem of 8 1 triangle The median line of a triangle is parallel to the third side and equal to half of it.
The trapezoid midline theorem is parallel to the two bases and equal to half of the sum of the two bases L = (a+b) ÷ 2s = l× h.
Basic properties of ratio 83 (1) If a:b=c:d, then ad=bc.
If ad=bc, then a:b=c:d wc ∕ /S∕?
84 (2) Combinatorial Properties If A/B = C/D, then (A B)/B = (C D)/D.
85 (3) Isometric Property If A/B = C/D = … = M/N (B+D+…+N ≠ 0), then
(a+c+…+m)/(b+d+…+n)=a/b
86 parallel lines are divided into segments and the theorem of proportionality. Three parallel lines cut two straight lines, and the corresponding segments are proportional.
It is inferred that the line parallel to one side of the triangle cuts the other two sides (or the extension lines of both sides), and the corresponding line segments are proportional.
Theorem 88 If the corresponding line segments obtained by cutting two sides (or extension lines of two sides) of a triangle are proportional, then this straight line is parallel to the third side of the triangle.
A straight line parallel to one side of a triangle and intersecting with the other two sides, the three sides of the cut triangle are directly proportional to the three sides of the original triangle.
Theorem 90 A straight line parallel to one side of a triangle intersects the other two sides (or extension lines of both sides), and the triangle formed is similar to the original triangle.
9 1 similar triangles's decision theorem 1 Two angles are equal and two triangles are similar (ASA)
Two right triangles divided by the height on the hypotenuse are similar to the original triangle.
Decision Theorem 2: Two sides are proportional and the included angle is equal, and two triangles are similar (SAS).
Decision Theorem 3 Three sides are proportional and two triangles are similar (SSS)
Theorem 95 If the hypotenuse and a right-angled side of a right-angled triangle are proportional to the hypotenuse and a right-angled side of another right-angled triangle, then the two right-angled triangles are similar.
96 Property Theorem 1 similar triangles corresponds to the height ratio, and the ratio corresponding to the median line and the ratio corresponding to the bisector are equal to the similarity ratio.
97 Property Theorem 2 The ratio of similar triangles perimeter is equal to similarity ratio.
98 Property Theorem 3 The ratio of similar triangles area is equal to the square of similarity ratio.
The sine of any acute angle is equal to the cosine of the remaining angles, and the cosine of any acute angle is equal to the sine of the remaining angles.
100 The tangent of any acute angle is equal to the cotangent of other angles, and the cotangent of any acute angle is equal to the tangent of other angles.
10 1 A circle is a set of points whose distance from a fixed point is equal to a fixed length.
102 The interior of a circle can be regarded as a collection of points whose center distance is less than the radius.
The outer circle of 103 circle can be regarded as a collection of points whose center distance is greater than the radius.
104 The radius of the same circle or equal circle is the same.
The distance from 105 to the fixed point is equal to the trajectory of a fixed-length point, which is a circle with the fixed point as the center and the fixed length as the half diameter.
106 and the locus of the point with the same distance between the two endpoints of the known line segment is the middle vertical line of the line segment.
The locus from 107 to a point with equal distance on both sides of a known angle is the bisector of this angle.
The trajectory from 108 to the equidistant point of two parallel lines is a straight line parallel and equidistant to these two parallel lines.
Theorem 109 Three points that are not on the same straight line determine a circle.
1 10 vertical diameter theorem divides the chord perpendicular to its diameter into two parts, and divides the two arcs opposite to the chord into two parts.
1 1 1 inference 1 ① bisect the diameter of the chord (not the diameter) perpendicular to the chord and bisect the two arcs opposite the chord.
(2) The perpendicular line of the chord passes through the center of the circle and bisects the two arcs opposite to the chord.
③ bisect the diameter of an arc opposite to the chord, bisect the chord vertically, and bisect another arc opposite to the chord.
1 12 Inference 2 The arcs sandwiched by two parallel chords of a circle are equal.
1 13 circle is a centrosymmetric figure with the center of the circle as the symmetry center.
Theorem 1 14 In the same circle or in the same circle, the isocentric angle has equal arc, chord and chord center distance.
1 15 It is inferred that in the same circle or equal circle, if one set of quantities in two central angles, two arcs, two chords or the chord-center distance between two chords is equal, the corresponding other set of quantities is also equal.
Theorem 1 16 The angle of an arc is equal to half its central angle.
1 17 Inference 1 The circumferential angles of the same arc or the same arc are equal; In the same circle or in the same circle, the arcs of equal circumferential angles are also equal.
1 18 Inference 2 The circumferential angle (or diameter) of a semicircle is a right angle; A chord with a circumferential angle of 90 is a diameter.
1 19 Inference 3 If the median line of one side of a triangle is equal to half of this side, then this triangle is a right triangle.
120 Theorem The diagonals of the inscribed quadrilateral of a circle are complementary, and any external angle is equal to its internal angle.
12 1① the intersection of the straight line l and ⊙O is d < r.
(2) the tangent of the straight line l, and ⊙ o d = r.
③ lines l and ⊙O are separated from each other, and d > r?
122 tangent theorem The straight line passing through the outer end of the radius and perpendicular to the radius is the tangent of the circle.
123 The property theorem of tangent line The tangent line of a circle is perpendicular to the radius passing through the tangent point.
124 Inference 1 A straight line passing through the center of the circle and perpendicular to the tangent must pass through the tangent point.
125 Inference 2 A straight line passing through the tangent and perpendicular to the tangent must pass through the center of the circle.
The tangent length theorem 126 leads to two tangents of a circle from a point outside the circle, and their tangent lengths are equal. The line between the center of the circle and this point bisects the included angle of the two tangents.
127 The sum of two opposite sides of a circle's circumscribed quadrilateral is equal.
128 Chord Angle Theorem The chord angle is equal to the circumferential angle of the arc pair it clamps.
129 Inference: If the arc enclosed by two chord tangent angles is equal, then the two chord tangent angles are also equal.
130 intersection chord theorem The length of two intersecting chords in a circle divided by the product of the intersection point is equal.
1365438+
132 tangent theorem leads to the tangent and secant of a circle from a point outside the circle, and the tangent length is the middle term in the length ratio of the two lines at the intersection of this point and secant.
133 It is inferred that two secant lines of the circle are drawn from a point outside the circle, and the product of the lengths of the two lines from that point to the intersection of each secant line and the circle is equal.
134 If two circles are tangent, then the tangent point must be on the line.
135① perimeter of two circles D > R+R ② perimeter of two circles d = r+r.
③ the intersection of two circles r-r < d < r+r (r > r)?
④ inscribed circle D = R-R (R > R) ⑤ two circles contain D < R-R (R > R).
Theorem 136 The intersection of two circles bisects the common chord of two circles vertically.
Theorem 137 divides a circle into n (n ≥ 3);
(1) The polygon obtained by connecting the points in turn is the inscribed regular N polygon of this circle.
(2) The tangent of a circle passing through each point, and the polygon whose vertex is the intersection of adjacent tangents is the circumscribed regular N polygon of the circle.
Theorem 138 Any regular polygon has a circumscribed circle and an inscribed circle, which are concentric circles.
139 every inner angle of a regular n-polygon is equal to (n-2) ×180/n.
140 Theorem Radius and apothem Divides a regular N-polygon into 2n congruent right triangles.
14 1 the area of the regular n polygon Sn = PNRN/2 P represents the perimeter of the regular n polygon.
142 The area of a regular triangle √ 3a/4a indicates the side length.
143 if there are k positive n corners around a vertex, since the sum of these angles should be 360, then K× (n-2) 180/n = 360 becomes (n-2)(k-2)=4.
144 arc length? Sword =n R/ 180
145 sector area formula: s sector =n r 2/360 = LR/2.
146 inner common tangent length = d-(R-r) outer common tangent length = d-(R+r)
There are still some, please help to supplement them. )
Practical tools: common mathematical formulas
Formula classification formula expression
Multiplication and factorization
a^2-b^2=(a+b)(a-b)
a^3+b^3=(a+b)(a^2-ab+b^2)
a^3-b^3=(a-b(a^2+ab+b^2)
Trigonometric inequality | A+B |≤| A |+B||||| A-B|≤| A |+B || A |≤ B < = > -b≤a≤b
|a-b|≥|a|-|b| -|a|≤a≤|a|
The solution of the unary quadratic equation -b+√ (b 2-4ac)/2a-b-√ (b 2-4ac)/2a
The relationship between root and coefficient x1+x2 =-b/ax1* x2 = c/a Note: Vieta theorem.
discriminant
B 2-4ac = 0 Note: The equation has two equal real roots.
b^2-4ac>； 0 Note: The equation has two unequal real roots?
b^2-4ac<； 0 Note: The equation has no real root, but a complex root of * yoke.
formulas of trigonometric functions
Two-angle sum formula
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-sinBcosA？
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/( 1-tanA tanB)
tan(A-B)=(tanA-tanB)/( 1+tanA tanB)
cot(A+B)=(cotA cotB- 1)/(cot B+cotA)？
cot(A-B)=(cotA cotB+ 1)/(cot b-cotA)
Double angle formula
tan2A=2tanA/[ 1-(tanA)^2]
cos2a=(cosa)^2-(sina)^2=2(cosa)^2 - 1= 1-2(sina)^2
half-angle formula
sin(A/2)=√(( 1-cosA)/2)sin(A/2)=-√(( 1-cosA)/2)
cos(A/2)=√(( 1+cosA)/2)cos(A/2)=-√(( 1+cosA)/2)
tan(A/2)=√(( 1-cosA)/(( 1+cosA))tan(A/2)=-√(( 1-cosA)/(( 1+cosA))
cot(A/2)=√(( 1+cosA)/(( 1-cosA))cot(A/2)=-√(( 1+cosA)/(( 1-cosA))？
Sum difference product
2sinAcosB=sin(A+B)+sin(A-B)
2cosAsinB=sin(A+B)-sin(A-B))
2cosAcosB=cos(A+B)-sin(A-B)
-2sinAsinB=cos(A+B)-cos(A-B)
sinA+sinB = 2 sin((A+B)/2)cos((A-B)/2
cosA+cosB = 2cos((A+B)/2)sin((A-B)/2)
tanA+tanB=sin(A+B)/cosAcosB
The sum of the first n terms of some series
1+2+3+4+5+6+7+8+9+…+n = n(n+ 1)/2
1+3+5+7+9+ 1 1+ 13+ 15+…+(2n- 1)= N2
2+4+6+8+ 10+ 12+ 14+…+(2n)= n(n+ 1)5
1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+…+n^2=n(n+ 1)(2n+ 1)/6
1^3+2^3+3^3+4^3+5^3+6^3+…n^3=n2(n+ 1)2/4
1 * 2+2 * 3+3 * 4+4 * 5+5 * 6+6 * 7+…+n(n+ 1)= n(n+ 1)(n+2)/3
Sine theorem a/sinA=b/sinB=c/sinC=2R Note: where r represents the radius of the circumscribed circle of a triangle.
Cosine Theorem B 2 = A 2+C 2-2 ACCOSB Note: Angle B is the included angle between side A and side C.
The standard equation of a circle (X-A) 2+(Y-B) 2 = R2 Note: (A, B) is the center coordinate.
General equation of circle x 2+y 2+dx+ey+f = 0 note: d 2+e 2-4f > 0.
Parabolic standard equation y 2 = 2px y 2 =-2px x 2 = 2py x 2 =-2py.
Lateral area of a straight prism S=c*h lateral area of an oblique prism s = c' * h.
Lateral area of a regular pyramid S= 1/2c*h' lateral area of a regular prism S= 1/2(c+c')h'
The lateral area of the frustum of a cone S = 1/2(c+c')l = pi(R+R)l The surface area of the ball S=4pi*r2.
Lateral area of cylinder S=c*h=2pi*h lateral area of cone s =1/2 * c * l = pi * r * l.
The arc length formula l=a*r a is the radian number r > of the central angle; 0 sector area formula s= 1/2*l*r
Cone volume formula V= 1/3*S*H cone volume formula V= 1/3*pi*r2h?
Oblique prism volume V=S'L Note: where s' is the straight cross-sectional area and l is the side length.
Cylinder volume formula V=s*h cylinder V=pi*r2h