2.(a+x)^m+ 1*(b+x)^n- 1-(a+x)^m*(b+x)^n
3.x^2+(a+ 1/a)xy+y^2
4.9a^2-4b^2+4bc-c^2
5.(c-a)^2-4(b-c)(a-b)
The answer is 1. The original formula = a4-a-3a+3 = (a-1) (a3+a2+a-3).
2.[ 1-(a+x)^m][(b+x)^n- 1]
3.(ax+y)( 1/ax+y)
4.9a^2-4b^2+4bc-c^2=(3a)^2-(4b^2-4bc+c^2)=(3a)^2-(2b-c)^2=(3a+2b-c)(3a-2b+c)
5.(c-a)^2-4(b-c)(a-b)
= (c-a)(c-a)-4(ab-b^2-ac+bc)
=c^2-2ac+a^2-4ab+4b^2+4ac-4bc
=c^2+a^2+4b^2-4ab+2ac-4bc
=(a-2b)^2+c^2-(2c)(a-2b)
=(a-2b-c)^2
1.x^2+2x-8
2.x^2+3x- 10
3.x^2-x-20
4.x^2+x-6
5.2x^2+5x-3
6.6x^2+4x-2
7.x^2-2x-3
8.x^2+6x+8
9.x^2-x- 12
10.x^2-7x+ 10
1 1.6x^2+x+2
12.4x^2+4x-3
Solve the equation: (x squared +5x-6) 1 =(x squared +x+6) 1.
Although cross multiplication is difficult to learn, it will bring us a lot of convenience once we learn it and use it to solve problems. The following is my personal opinion on cross multiplication.
1, the method of cross multiplication: the left multiplication of the cross is equal to the quadratic term coefficient, the right multiplication is equal to the constant term, and the cross multiplication is equal to the linear term coefficient.
2. Use of cross multiplication: (1) Cross multiplication is used to decompose factors. (2) Solving the quadratic equation with one variable by cross multiplication.
Advantages of cross multiplication: it is faster to solve problems with cross multiplication, which can save time and is not easy to make mistakes.
4. Defects of cross multiplication: 1. Some problems are relatively simple to solve by cross multiplication, but not every problem is simply solved by cross multiplication. 2. Cross multiplication is only applicable to quadratic trinomial type problems. 3. Cross multiplication is more difficult to learn.
5. Examples of cross multiplication problem solving:
1), using cross multiplication to solve some simple and common problems.
Example 1 M? +4m- 12 factorization factor
Analysis: The constant term-12 in this question can be divided into-1× 12, -2×6, -3×4, -6×2,-12× 1 2.
Solution: Because 1 -2
1 ╳ 6
So m? +4m- 12=(m-2)(m+6)
Example 2 Handle 5x? +6x-8 factorization factor
Analysis: In this question, 5 can be divided into 1× 5, and -8 can be divided into-1×8, -2×4, -4×2, -8× 1. When the coefficient of quadratic term is divided into 1×5 and the constant term is divided into -4×2, it is consistent with this question.
Solution: Because 1 2
5 ╳ -4
So 5x? +6x-8=(x+2)(5x-4)
Example 3 solving equation x? -8x+ 15=0
Analysis: put x? -8x+ 15 is regarded as a quadratic trinomial about x, then 15 can be divided into 1× 15 and 3×5.
Solution: Because 1 -3
1 ╳ -5
So the original equation can be transformed into (x-3)(x-5)=0.
So x 1=3 x2=5.
Example 4. Solve equation 6x? -5x-25=0
Analysis: put 6x? If -5x-25 is regarded as a quadratic trinomial about X, then 6 can be divided into 1×6, 2×3 and -25 can be divided into-1×25, -5×5 and -25× 1.
Solution: Because 2 -5
3 ╳ 5
So the original equation can be changed to (2x-5)(3x+5)=0.
So x 1=5/2 x2=-5/3.
2) Use cross multiplication to solve some difficult problems.
Example 5 14x? -67xy+ 18y？ Decomposition factor
Analysis: put 14x? -67xy+ 18y？ As a quadratic trinomial about X, 14 can be divided into 1× 14, 2×7, 18y? It can be divided into y. 18y, 2y.9y and 3y.6y
Solution: Because 2-9 years old
7 ╳ -2y
So 14x? -67xy+ 18y？ =(2 to 9 years) (7 to 2 years)
Example 6 10x? -27xy-28y？ -x+25y-3 factorization factor
Analysis: This question should organize this polynomial into a quadratic trinomial form.
Solution 1, 10x? -27xy-28y？ -x+25y-3
= 10x？ -(27y+ 1)x -(28y？ -25y+3) 4y -3
7y ╳ - 1
= 10x？ -(27y+ 1)x-(4y-3)(7y- 1)
=[2x-(7y- 1)][5x+(4y-3)]2-(7y– 1)
5 ╳ 4y - 3
=(2x -7y + 1)(5x +4y -3)
Note: this question, first put 28y? -25y+3 is decomposed into (4y-3)(7y-1), 10x? -(27y+1) x-(4y-3) (7y-1) is decomposed into [2x -(7y-1)][5x +(4y -3)].
Solution 2, 10x? -27xy-28y？ -x+25y-3
=(2x-7y)(5x+4y)-(x-25y)-3 ^ 2-7y
=[(2x-7y)+ 1][(5x-4y)-3]5╳4y
=(2x-7y+ 1)(5x-4y-3)2x-7y 1
5 x - 4y ╳ -3
Note: For this question, put 10x first. -27xy-28y？ It is decomposed into (2x -7y)(5x +4y) by cross multiplication, and then it is decomposed into [(2x -7y)+ 1] [(5x -4y)-3] by cross multiplication.
Example 7: Solve the equation about x: x? - 3ax + 2a？ -a B- b？ =0
Analysis: 2a? -a B- b？ Cross multiplication can be used for factorization.
Solution: x? - 3ax + 2a？ -a B- b？ =0
x？ - 3ax +(2a？ -a B- b？ )=0
x？ - 3ax +(2a+b)(a-b)=0 1 -b
2 ╳ +b
[x-(2a+b)][x-(a-b)]= 0 1-(2a+b)
1 ╳ -(a-b)
So x1= 2a+bx2 = a-b.
5-7(a+ 1)-6(a+ 1)^2
=-[6(a+ 1)^2+7(a+ 1)-5]
=-[2(a+ 1)- 1][3(a+ 1)+5]
=-(2a+ 1)(3a+8)；
-4x^3 +6x^2 -2x
=-2x(2x^2-3x+ 1)
=-2x(x- 1)(2x- 1)；
6(y-z)^2 + 13(z-y)+6
=6(z-y)^2+ 13(z-y)+6
=[2(z-y)+3][3(z-y)+2]
=(2z-2y+3)(3z-3y+2)。
For example, the formula ... x 2+6x-7
Because the coefficient before the first power x is 6.
Therefore, we can consider 7- 1=6.
That's just the constant term of this formula is -7.
So we think of -7 as 7 *( 1).
So we cross each other.
x +7
x - 1
To (x+7) (x- 1)
The factor was successfully decomposed.
3ab^2-9a^2b^2+6a^3b^2
=3ab^2( 1-3a+2a^2)
=3ab^2(2a^2-3a+ 1)
=3ab^2(2a- 1)(a- 1)
5-7(a+ 1)-6(a+ 1)^2
=-[6(a+ 1)^2+7(a+ 1)-5]
=-[2(a+ 1)- 1][3(a+ 1)+5]
=-(2a+ 1)(3a+8)；
-4x^3 +6x^2 -2x
=-2x(2x^2-3x+ 1)
=-2x(x- 1)(2x- 1)；
6(y-z)^2 + 13(z-y)+6
=6(z-y)^2+ 13(z-y)+6
=[2(z-y)+3][3(z-y)+2]
=(2z-2y+3)(3z-3y+2)。
For example, the formula ... x 2+6x-7.
Because the coefficient before the first power x is 6.
Therefore, we can consider 7- 1=6.
That's just the constant term of this formula is -7.
So we think of -7 as 7 *( 1).
So we cross each other.
x +7
x - 1
To (x+7) (x- 1)
The factor was successfully decomposed.
3ab^2-9a^2b^2+6a^3b^2
=3ab^2( 1-3a+2a^2)
=3ab^2(2a^2-3a+ 1)
=3ab^2(2a- 1)(a- 1)
x^2+3x-40
=x^2+3x+2.25-42.25
=(x+ 1.5)^2-(6.5)^2
=(x+8)(x-5)。
[6] Cross multiplication.
There are two situations in this method.
① factorization of x2+(p+q) x+pq formula.
The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).
② Factorization of KX2+MX+N formula
If k=ac, n=bd, ad+bc=m, then kx 2+MX+n = (ax+b) (CX+d).
The chart is as follows:
A b
×
c d
For example, because
1 -3
×
7 2
-3× 7 =-2 1, 1× 2 = 2, and 2-2 1=- 19,
So 7x 2- 19x-6 = (7x+2) (x-3).
Formula of cross multiplication: head-tail decomposition, cross multiplication and summation.
⑶ Grouping decomposition method
Group decomposition is a simple method to solve equations. Let's learn this knowledge.
There are four or more terms in an equation that can be grouped, and there are two forms of general grouping decomposition: dichotomy and trisection.
For example:
ax+ay+bx+by
=a(x+y)+b(x+y)
=(a+b)(x+y)
We put ax and ay in a group, bx and by in a group, and matched each other by multiplication and division and distribution, which immediately solved the difficulty.
Similarly, this problem can be done.
ax+ay+bx+by
=x(a+b)+y(a+b)
=(a+b)(x+y)
A few examples:
1.5ax+5bx+3ay+3by
Solution: =5x(a+b)+3y(a+b)
=(5x+3y)(a+b)
Note: Different coefficients can be decomposed into groups. As mentioned above, 5ax and 5bx are regarded as a whole, and 3ay and 3by are regarded as a whole, which can be easily solved by using the multiplication and distribution law.
2.x3-x2+x- 1
Solution: =(x3-x2)+(x- 1)
=x2(x- 1)+(x- 1)
=(x- 1)(x2+ 1)
Using dichotomy, the common factor method is used to propose x2, and then it is easy to solve it.
3.x2-x-y2-y
Solution: =(x2-y2)-(x+y)
=(x+y)(x-y)-(x+y)
=(x+y)(x-y+ 1)
Use dichotomy, then use the formula a2-b2=(a+b)(a-b), and then skillfully solve it.
758? —258? =(758+258)(758-258)= 10 16*500=508000