[2x + a] * [3x + b] one person wrongly copies a in the first polynomial to get the square of result 6x plus 11x-10. Another person wrongly copies the coefficient of X in the second monomial to get the square of result 2x - 9x + 10. Find the value of a and B

[2x + a] * [3x + b] one person wrongly copies a in the first polynomial to get the square of result 6x plus 11x-10. Another person wrongly copies the coefficient of X in the second monomial to get the square of result 2x - 9x + 10. Find the value of a and B

a=-5,b=-2
[2x + a] * [3x + b] = the square of 6x + (3a + 2b) x + ab
From the constant terms of the first and the second, we can see that ab = 10. The mistake is to copy a into - A, that is - 3A + 2B = 11
From the coefficients of the square terms of X of the first and the second, we can see that the error coefficient is: copy 3 to 1, that is, a + 2B = - 9
Finally, a = - 5 and B = - 2 are obtained from the solution of (1) and (2)
Party A and Party B calculate an integral multiplication problem together: (2x + a) (3x + b). Because party a miscopies the sign of a in the first polynomial, the result is 6x + 11x-10; because Party B omits to copy the coefficient of X in the second polynomial, the result is 2x ^ 2 - 9x + 10
Find 1: what are the values of a and B in the formula
2: Calculate the correct result of this integral multiplication problem
According to the result (2x-a) (3x + b) = 6x ^ 2-3ax + 2bx AB = 6x ^ 2 - (3a-2b) x-ab according to the meaning - (3a-2b) = 11 (1) - AB = - 10 according to the result (2x + a) (x + b) = 2x ^ 2 + ax + 2bx + AB = 2x ^ 2 + (a + 2b) x + ABA + 2B = - 9 (2) AB = 103a-2b = - 11 (1) a + 2B = - 9 (2) (1) + (
Analysis,
The formula obtained by a: (2x-a) (3x + b) = 6x & # 178; + (2b-3a) x-ab = 6x & # 178; + 11x-10
The corresponding coefficients are equal, 2b-3a = 11, and ab = 10
The formula of B: (2x + a) (x + b) = 2x & # 178; + (2B + a) x + AB = 2x & # 178; - 9x + 10
The corresponding coefficients are equal, 2b + a = - 9, and ab = 10
2b-3a = 11, and 2b + a = - 9
Analysis,
The formula obtained by a: (2x-a) (3x + b) = 6x & # 178; + (2b-3a) x-ab = 6x & # 178; + 11x-10
The corresponding coefficients are equal, 2b-3a = 11, and ab = 10
The formula of B: (2x + a) (x + b) = 2x & # 178; + (2B + a) x + AB = 2x & # 178; - 9x + 10
The corresponding coefficients are equal, 2b + a = - 9, and ab = 10
2b-3a = 11, and 2b + a = - 9,
a=-5,b=-2
Correct formula: (2x-5) (3x-2) = 6x & # 178; - 19x + 10
∵ A is given as follows: (2x-a) (3x + b) = 6x2 + (2b-3a) x-ab = 6x2 + 11x-10
The corresponding coefficients are equal, 2b-3a = 11, ab = 10,
The formula of B: (2x + a) (x + b) = 2x2 + (2B + a) x + AB = 2x2-9x + 10
The corresponding coefficients are equal, 2b + a = - 9, ab = 10,
Qi
2b-3a=112b+a=-9，
The solution is as follows
A. unfold
∵ A is given as follows: (2x-a) (3x + b) = 6x2 + (2b-3a) x-ab = 6x2 + 11x-10
The corresponding coefficients are equal, 2b-3a = 11, ab = 10,
The formula of B: (2x + a) (x + b) = 2x2 + (2B + a) x + AB = 2x2-9x + 10
The corresponding coefficients are equal, 2b + a = - 9, ab = 10,
Qi
2b-3a=112b+a=-9，
The solution is as follows
a=-5b=-2，
The correct formula: (2x-5) (3x-2) = 6x2-19x + 10
It is known that: (2x-a) (3x + b) = 6x ^ 2 + (2b-3a) x-ab = 6x ^ 2 + 11x-10, so 2b-3a = 11, ab = 10
And (2x + a) (x + b) = 2x ^ 2 + (a + 2b) x + AB = 2x ^ 2-9x + 10, so a + 2B = - 9, ab = 10, we can get a = - 5, B = - 2,
So the original formula = (2x-5) (3x-2) = 6x ^ 2-19x + 10
A and B calculate the integral multiplication: (2x + a) (3x + b) together. Because a miscopies the sign of a in the first polynomial, the result is 6x ^ 2 + 11x-10; because B omits to copy the coefficient of X in the second polynomial, the result is 2x ^ 2-9x + 10
Can you find out the values of a and B in the formula?
A: (2x-a) (3x + b) = 6x & # 178; + 11x-10, let x = 0, then - AB = - 10, ab = 10
B: (2x + a) (x + b) = 2x & # 178; - 9x + 10, let x = 1, we get (2 + a) (1 + b) = 3,2 + 2B + A + AB = 3,2b + a = - 9, we get a = - 5, B = - 2, or a = - 4, B = - 2.5
If the fractional equation 1 + 3K / x + 1 = k has no solution, then the value of K is
1+3k=kx+k
KX = - 2k-1, if there is no solution, then k = 0
1+3k=kx+k
kx=-1-2k
(1) K = 0, the equation has no solution
(2) When x = - 1, i.e. k = 1, the equation has no solution.
It is known that the triangle a, B and C satisfy the following conditions: the square of a + the square of B + the square of C = AB + BC + AC
Multiply both sides of the equation by two
2A ^ 2 + 2B ^ 2 + 2C ^ 2 = 2Ab + 2BC + 2Ac (a ^ 2 is the square of a)
2a^2+2b^2+2c^2-(2ab+2bc+2ac)=0
a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0
(a-b)^2+(b-c)^2+(c-a)^2=0
So (a-b) ^ 2 = 0 (B-C) ^ 2 = 0 (C-A) ^ 2 = 0
So a = B, B = C, C = a
So a = b = C
The triangle ABC is an equilateral triangle
When k is a value, the equation X-1 / 3 (2x + k) + 3K = 1-1 / 2 (x-3k) about X is a process of finding positive numbers
x-1/3（2x+k)+3k=1-1/2（x-3k）
x-2x/3-k/3+3k=1-x/2+3k/2
5x/6=1-7k/6
x=(6-7k)/5
Let X be greater than 0
That is, (6-7K) / 5 > 0
Get k
If a2-ab = 8 and ab-b2 = - 4, then A2-B2=______ ，a2-2ab+b2=______ .
A2-ab = 8, ①, ab-b2 = - 4, ②, ① + ②: A2-B2 = 4; ① - ②: a2-2ab + B2 = 12
Solving fractional equation (x-4) / (X-5) - (X-5) / (X-6) = (X-7) / (X-8) - (X-8) / (X-9)
Multiply left and right sides by (X-5) (X-6) (X-8) (X-9)
(x-4) (X-6) (X-8) (X-9) - (X-5) (X-5) (X-8) (X-9) = (X-5) (X-6) (X-7) (X-9) - (X-5) (X-6) (X-8) (X-8)
(X-8) (X-9) = (X-5) (X-6)
The expanded solution is x = 7
Let y = X-5, then the left side of the equation is (y + 1) / Y-Y / (Y-1) = - 1 / (Y & # 178; - y) = - 1 / (X & # 178; - 11x + 30)
Color z = X-8, then the right side of the equation is (Z + 1) / Z-Z / (Z-1) = - 1 / (Z & # 178; - z) = - 1 / (X & # 178; - 17x + 72)
Left = right
-1/(x²-11x+30)=-1/(x²-17x+72)
That is, X & # 178; - 11x + 30 = x & # 178; - 17x + 72
We get x = 7.
If the square of a-Ab = 9 and the square of ab-b = 6, then what is the square of A-B and the square of a-2ab + B
Root 15,15
Fractional equation. X / (X-2) + (X-9) / (X-7) = (x + 1) / (x-1) + (X-8) / (X-6)
(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/[(x-6)(x-2)][x^2-10x+9-x^2+6x+7]/(x^2-8x+7]=(x^2-10x+16-x^2+6x]/[x^2-8x+12](-4x+16)/(x^2-8x+7)=(-4x+...