Xiaoqing and Xiaohong calculate the same integral multiplication problem: (2x + a) (3x + b) respectively. Xiaoqing gets 6x2-13x + 6 by copying the wrong sign of a in a polynomial, and Xiaohong gets 2x2-x-6 by copying the wrong coefficient of X in the second polynomial. Then the correct result of this problem is

Xiaohong's answer is 2x & # 178; - X-6, so her formula is: (2x + a) (x + b) = 2x & # 178; + (a + 2b) + AB = 2x & # 178; - X-6 # a + 2B = - 1, ab = - 6, Xiaoqing's formula is: (2x-a) (3x + b) = 6x & # 178; + (- 3A + 2b) - AB = 6x & # 178; - 13X + 6 # 3A + 2B = - 13, ② AB = - 6
Qingqing and Yuyu calculate an integral multiplication (2x + a) × (3x + b) together. Because Qingqing miscopied the sign of a in the first polynomial, the result 6a2-13x + 6 is obtained. Because Yuyu missed the coefficient of X in the second polynomial, the result is 2x2-x-6. Find the value of a, B and the correct result of this integral
The square of a multiplied by the square of a of b-ab-the square of B divided by (1 + 2Ab the square of a + the square of B)
(a-h-178; a-b-b-b-ab-ab-b-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-ab-178; (; (a) (a-a-178; (a-h-178; (a-178; (a-a-h-178; (a-h-178; (b-b-b-b-b-178; (b-b-b-b-b-b-b-b-b-b-b-b-b-b-b-ab] = [(a (a-a-a-b) (a-a-a-b) (a-a-a-b) (a-178; (a-b-b-b-b-b-b-b-b-178; (b-178; (b) (b-b-b-b-b-b-b-b-178; (b-b-b-b-b-b-b-b-b-b) / (a + b) &
Solve the equation 7X2 + X − 3x − x2 = 1 + 7 − x2x2 − 1
The original equation can be reduced to: 7x (x + 1) + 3x (x − 1) = 1 + 7 − X2 (x + 1) (x − 1) by multiplying x (x + 1) (x-1) on both sides of the equation, 7 (x-1) + 3 (x + 1) = x (x + 1) (x-1) + X (7-x2) simplified, 4x = 4  x = 1 test: substituting x = 1 into x (x + 1) (x-1) = 0  x = 1 is the increasing root of the original equation
If the square of a-2ab + the square of B = 0, then the square of A-A of b-ab of B + the square of a + the square of B
Square of a-2ab + square of B = 0
∴(a-b)²=0
∴a=b
A-a-b-ab the square of a + B
=B / A-B / B - (B & # 178; + B & # 178;) / B & # 178; replace a with B
=-2
Solving the fractional equation 200 / (6 + x) - 200 / (8 + x) = 5
All of a sudden, I was in a hurry
200/(6+x)-200/(8+x)=5
200(x+8)-200（x+6）=5(x+6)(x+8);
400=5(x²+14x+48);
x²+14x+48=80;
x²+14x-32=0;
(x+16)(x-2)=0;
X = 2 or x = - 16;
X=2！！！！！！！！！！！！！！！！！！！！
A squared B + AB squared a squared-b squared divided by (2 a squared-b squared-1)
Binary linear fractional equation (1) x + Y / 4 + X-Y / 6 = 3 X-Y / 9-x + Y / 1 = 1 (mainly process)
Let x + y = a, X-Y = B
So 4 / A + 6 / b = 3
-1 / A + 9 / b = 1, so - 4 / A + 36 / b = 4
Add it up to: 42 / b = 7, so B = 6, a = 2
The equation 3x-2y = 15,5x-4y = 23 is solved by the method of substitution
3x-2y=15
-2y=15-3x
Substituting 2
5X+2(15-3X)=23
5X+30-6X=23
X=7
Y=3
Angle 1 plus angle 2 plus angle 3 plus angle 4 is 180 degrees. The answer is five pairs of complementary angles, but I don't know which five pairs?
Solving the fractional equation: 4 / 5x + X / (5x-25) = 5 / (x ^ 2-5x)
4 / 5x + X / (5x-25) = 5 / (x ^ 2-5x) (4x-20) / 5x (X-5) + x ^ 2 / 5x (X-5) = 25 / 5x (X-5) 4x-20 + x ^ 2 = 254x + x ^ 2 = 45x ^ 2 + 4x + 2 ^ 2 = 45 (x + 2) ^ 2 = 45 + 2 ^ 2x + 2 = positive and negative root 49x = - 9 or 5, because in X / (5x-25), when x = 5, 5x-25 = 0 (denominator cannot be 0), so x = 5 does not conform to the meaning of the question
Is it X / (X-2) - (1-x ^ 2) / (x ^ 2-5x + 6) = 2x / (x-3)? First of all, factorization (x ^ 2-5x + 6) = (X-2) (x-3), it is obvious that x is not equal to 2 and X is not equal to 3, so the following question: no

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