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Mathematics calculation problems in the first volume of the second day of junior high school
There are 200 math problems in senior two. If it is a little harder, there is no answer: Wen Ku. Baidu, please. /view/e 9929 a3 d 5727 a5e 9856 a 6 1 E4。 This should do.

Junior high school math problems, the more difficult it is to buy a book "three-year simulated five-year senior high school entrance examination" is good for you!

100 A slightly more difficult mathematical decimal calculation problem is urgently needed! (write the answer! ) come to the following website.

Find 10 problem (with answer) 1, (3ab-2a) ÷÷÷.

2、(x^3-2x^y)÷(-x^2)

3、-2 1a^2b^3÷7a^2b

4、(6a^3b-9a^c)÷3a^2

5、(5ax^2+ 15x)÷5x

6 、( a+2b)(a-2b)

7、(3a+b)^2

8 、( 1/2 a- 1/3 b)^2

9 、( x+5y)(x-7y)

10 、( 2a+3b)(2a+3b)

Answer:

1 、( 3ab-2a)÷a =3b-2

2、(x^3-2x^y)÷(-x^2) =-x+2x^(y-2)[? ]

3、-2 1(a^2)(b^3)÷7(a^2)b =-3b^2

4、(6(a^3)b-9a^c)÷3a^2 =2ab-3a^(c-2)[? ]

5、(5ax^2+ 15x)÷5x =ax+3

6 、( a+2b)(a-2b) =a^2-4b^2

7、(3a+b)^2 =9a^2+6ab+b^2

8 、( 1/2 a- 1/3 b)^2 = 1/4a^2- 1/3ab+ 1/9b^2

9. (x+5 years) (x-7 years) = x 2-2xy-12y 2.

10 、( 2a+3b) 4a^2+ 12ab+9b^2

The process of solving five difficult problems in junior one mathematics is 1. Calculation: square of 20072005+square of 20072007-2/square of 20072006 =?

2. It is known that A2B2+A2+B2+ 1=4AB. Find the values of a and b (for simplicity, the square of a is omitted as A2. Except 4AB, the others are the same as A2. )

3. Assuming that A, B and C are three sides of a triangle, and A2+B2+C2-AB-BC-CA=0, try to explain that a triangle is an equilateral triangle. (same as above)

4 A/|A|+B/|B|+C/|C|= 1。 Find the value of (| ABC |/ABC) to the power of 2007 (| AB |/BC+| BC |/AC+|/AB).

answer

1, let 20072006=x, then

(x- 1)^2+(x+ 1)^2-2x^2=x^2-2x+ 1+x^2+2x+ 1-2x^2=2

2、A^2B^2+A^2+B^2+ 1-4AB=0

(ab- 1)^2+(a-b)^2=0 = & gt; AB= 1,A = B = & gtA=B=0.5

3、a^2+b^2+c^2-ab-bc-ca= 1/2[(a-b)^2+(b-c)^2+(c-a)^2]=0

=>A = B = C => is an equilateral triangle.

4. There are two negative numbers and one positive number in A/|A|+B/|B|+C/|C|= 1 = >A, B and C, so

ABC =-| ABC | = & gt; |ABC|/ABC=- 1

The answer to the math problem in grade two. . . ( 1)、9x(4x-7)-(6x+5)(6x-5)+38=0

36x^2-63x-36x^2+25+38=0

63x=63

x= 1

(2)、-3y(y^2-2y+ 1)+y^2(3y-6)≤6

-3y^3+6y^2-3y+3y^3-6y^2≤6

-3y≤6

y≥-2

Find the answers to two math problems in grade two, 1, s/(a+b)-s/(a-b) = s (a-b)-s (a+b)/(a-b) = as-bs-as-bs/a2.

2、250/a+250/(20-a)=250*(20-a)+250a/a*(20-a)=5000-250a+250a/20a-a^2=5000-20a-a^2

Note:/indicates fractional line.

Do you have any math problems? Junior high school, more difficult! Three people went to the hotel and stayed for 30 yuan a night. Nobody gave the boss 10 yuan, and then the boss said that today's discount is only 25 yuan. Ask the waiter to get back 5 yuan for them. The waiter secretly hid 2 yuan, and each person got 1 yuan. 9x3=27, and each of them only paid 9 yuan in the end. 27+ waiter 2 yuan =29, and where did 1 yuan go? Give the answer o(∩_∩)o ... After adoption.

Mathematics calculation problems and answers in the mid-term exam of junior two mathematics

Class name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

I. Selection (3 points for each small question *** 10 small question)

1. The following statement is incorrect ()

The center of a triangle is the intersection of three bisectors of the triangle.

B the point where the three vertices of a triangle are equidistant is the intersection of the perpendicular lines of the three sides.

Of the three internal angles of any triangle, there are at least two acute angles.

D. Two right-angled triangles with a common hypotenuse are congruent.

2. If all three sides of a triangle are integers, the perimeter is 1 1, and one side is 4, then the longest side in this triangle is ().

a7 b . 6 c . 5d . 4

3. Factorization is ()

A.B.

C.D.

4.a and B are rational numbers of (a≠b), and the value of is ().

A.B. 1 C.2 D.4

5. If the angle between the height and the bottom of an isosceles triangle is 45, then the triangle is ().

A. acute triangle B. obtuse triangle C. equilateral triangle D. isosceles right triangle

6. It is known that x should satisfy ()

A.x < 2 b.x ≤ 0 c.x > 2 d.x ≥ 0 and x≠2

7. As shown in the figure, in △AB=AC, AB = AC, DE is the middle vertical line of AB side, the circumference of △BEC is 14cm, and BC = 5cm, then the length of AB is ().

a . 14cm b . 9cm c . 19cm d . 1 1cm

8. The following calculation is correct ()

A.B.

C.D.

9. If ... is known, the value of is ()

A. 15

10. There are four propositions, of which the correct one is ().

(1) congruence of an isosceles triangle with an angle of 100.

(2) Among the straight lines connecting two points, the straight line is the shortest.

(3) A triangle with two equal angles is an isosceles triangle.

(4) In △ABC, if ∠ A-∠A-∠B = 90°, △ABC is an obtuse triangle.

A.( 1)(2)b .(2)(3)c .(3)(4)d .( 1)(4)

Fill in the blanks (2 points for each small question *** 10 small question)

1. If known, then _ _ _ _ _ _ _ _ _ _ _ _

2. Decomposition coefficient _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

3. When x = _ _ _ _ _ _ _ _ _ _ _ _ _ _ time division value is zero.

4. If X = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

5. Calculate _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

6. The circumference of an isosceles triangle is = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

7. If the outer angle of the top angle of an isosceles triangle is 30 smaller than the outer angle of the bottom angle, then the inner angle of this triangle is _ _ _ _ _ _ _.

_____________________

8. As shown in △ABC, if AD⊥BC is in D, ∠ B = 30, ∠ C = 45, CD = 1, then AB = _ _ _ _ _ _ _ _ _

9. As shown in △ABC, BD bisector ∠ABC, BD⊥AC at D, DE‖BC, AB at E, AB = 5cm, AC = 2cm, then the circumference of △ADE = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

10. In △ABC, ∠ c = 1 17, the perpendicular line on the side of AB intersects BC in D, and AD is divided into ∠CAB. ∠ CAD: ∠ DAB = 3: 2, then ∠.

Third, the calculation problem (***5 small questions)

1. decomposition (5 points)

2. Calculation (5 points)

3. Simplify and re-evaluate, where x =-2 (5 points)

4. Solve the equation (5 points)

In order to alleviate the traffic jam, it is decided to build a light rail from the city center to the airport. In order to finish the project three months ahead of schedule, it is necessary to improve the original planned work efficiency by 12%. How many months will it take to plan this project? (6 points)

Fourth, proof calculation and drawing (***4 small questions)

1. As shown in the figure, when △AB=AC, AB=AC, ∠ A = 120, DF divides AB, AB, F and BC vertically, D. Verification: (5 points)

2. As shown in Figure C, a point on AB is an equilateral triangle with △AMC and △CNB. Verify that An = BM (6 points).

3. Find a point P so that PC = PD and the distance from point P to ∠AOB is equal. (No writing method) (5 points)

4. As shown in the line segment BD of points E and F, AB = CD, ∠ B = ∠ D, BF = de. (8 points).

Verification (1) AE = cf

(2)AE‖CF

(3)AFE =∠CEF

Reference answer

I. Selection (3 points for each small question *** 10 small question)

1.D 2。 C 3。 D 4。 B 5。 D 6。 B 7。 B 8。 C 9。 B 10。 C

Fill in the blanks (2 points for each small question *** 10 small question)

1.2 2.3. 1 4.5 5.

6.7 7.80 50 50 8.2 9.7 cm 10. 18

Third, the calculation problem (***5 small questions)

1. solution:

2. Solution:

.

3. Solution:

party history

The value of the original formula.

4. Solution:

.

Test: X = 4 is the root of the original equation.

It was originally planned that this project would take x months.

Test is the root of the original equation.

A: It was originally planned to be completed in 28 months.

Fourth, proof calculation and drawing (***4 small questions)

1. Certificate: even AD.

∫∠A = 120

AB=AC

∴ ∠B=∠C=30

∫fd⊥ Divide AB equally.

∴ BD=AD

∠B=∠ 1=30

∠DAC=90

∫In Rt△ADC

∠C=30

that is

2. Certificate: Point ∵ C is on AB.

A, b and c are in a straight line.

∠ 1+∠3+∠2= 180

∵ △AMC and△△△ CNB are equilateral triangles.

∴ ∠ 1=∠2=60

That is ∠ 3 = 60.

AC=MC,

CN=CB

In △MCB and △ACN.

∴△MCB?△ACN(SAS)

∴ Ann =MB.

3.

4. Syndrome ① is in △ABF and △DCE.

∴△abf?△DCE(SAS)

∴ AF=CE,∠ 1=∠2

B, f, e and d are in a straight line.

∴∠ 3 = ∞∞∠∠ 4 (complementary angles of the same angle are equal)

That is ∠ AFE = ∠ cef.

② in △AFE and △CEF.

∴△AFE?△cef(SAS)

∴ AE=CF ∠5=∠6

∵ ∠5=∠6

∴ AE‖CF

③ ∵ ∠3=∠4

That is, ∠ AFE = ∠ cef.

I want a slightly more difficult junior high school math problem. It is known that the solution of equation a-x/2=bx/3 about X is x=2, where a≠0 and b≠0. Find the value of algebraic expression a/b-b/a.

If you can't find any difficulty, take advantage of it.