It is known that a (square of X + x-C) + B (square of 2x-x-2) = square of 7x + 4x + 3 What are the values of a, B and C

It is known that a (square of X + x-C) + B (square of 2x-x-2) = square of 7x + 4x + 3 What are the values of a, B and C

Left = ax? Ax AC + 2bx? - bx-2b
=(a+2b)x²+(a-b)x+(-ac-2b)
=Right = 7x? + 4x + 3
So the corresponding coefficients are equal
therefore
a+2b=7
a-b=4
-ac-2b=3
So a = 5, B = 1, C = - 1

Minus 6x square plus 7x + 5 = (2x + 1) times a, then a =?

-(2*x+1)*(3*x-5)=(2*x+1)*A
So when 2 * x + 1 ≠ 0, that is, X ≠ - 1 / 2
A=-3*x+5
When x = - 1 / 2, a is any value

Given that a = - 2x + 3, B = - 7x squared - 6x + 5, find 2a-3b

We know that a = - 2x + 3, B = - 7x squared - 6x + 5
What does this mean? It means that the square of B equals - 6x + 5

When x = - 2, calculate the square of 7x-3x the square of-2x-2x + 5 + 6x

Five

If two equations a ^ 2x ^ 2 + AX-1 = 0 and x ^ 2-ax-a ^ 2 = 0 have a common heel, find the value of A A ^ 2 and x ^ 2 are separate. Who can help

If the common root of the equation is B, then it is substituted into the above two equations
(ab)^2+ab-1=0
b^2-ab-a^2=0
Add the above two equations:
---->b^2(a^2+1)-(a^2+1)=0
---->(b^2-1)(a^2+1)=0
---->B = 1 or - 1;
1. When B = 1, it is substituted into the second equation: A ^ 2 + A-1 = 0; according to the root formula, a = (- 1 ±√ 5) / 2 can be obtained
2. When B = - 1, put it into the second equation: A ^ 2-a-1 = 0; according to the root formula, a = (1 ± √ 5) / 2 can be obtained

On the equation ax-2x = 6 of X (1) through the introduction test, it is found that when a = 0, the solution of the equation is x = - 3; when a = 1, the solution of the equation is x = - 6; when a = 2, the equation has no solution; when a = 3, the solution of the equation is x = 6. What is the relationship between the solution of the equation and the value of a? (2) When a is the value, the equation has a positive integer solution? And find the positive integer solution (3) When a is the value of a, the equation has an integer solution

For the equation ax-2x = 6
(a-2)x=6
（1）
① When a = 2, the equation has no solution;
② When a ≠ 2, the solution of the equation is x = 6 / (A-2)
(2) If the equation has a positive integer solution, then a ≠ 2, and 6 / (A-2) is a positive integer,
That is, A-2 is a positive divisor of 6, and the positive divisor of 6 has 6,3,2,1,
When a is 3,4,5,8,
The integral solutions of the equation are 6,3,2,1
(3) If the equation has an integer solution, then a ≠ 2 and 6 / (A-2) is an integer,
That is, A-2 is the divisor of 6, and the divisor of 6 is ± 6, ± 3, ± 2, ± 1,
When a is - 4, - 1,0,1,3,4,5,8,
The integral solutions of the equation are - 1, - 2, - 3, - 6,6,3,2,1, respectively

If the inequality system x > 1 / 2 (x-3), the integer solution of 2x + 3 < 1 is the root of the equation 2X-4 = ax about X, then find the value of A

First, two inequalities are solved
X 〉 - 3 (first formula)
X 〈 - 1 (second formula)
In the range of - 3 to - 1, only one integer can be taken, which is - 2
The root of 2X-4 = ax is x = - 2
Substitute: - 2A = - 8
So a = 4
Come on!

If the smallest integer solution of inequality 5 (X-2) + 8 < 6 (x-1) + 7 is the solution of equation 2x AX = 3, find the value of A

5（x-2）+8<6(x-1)+7
5x-10+8<6x-6+7
5x-6x<-6+7+10-8
-x<3
x>-3
The minimum integer solution of 5 (X-2) + 8 < 6 (x-1) + 7 is - 2
Taking x = - 2 generation equation 2x-3a = 3, we get the following results:
-4-3a=3
-3a=7
a=-7/3

(1) Solving inequality: 5 (X-2) + 8

（1）5x-10+8< 6x-6+7
5x-2 -3
(2) The smallest integer satisfying x > - 3 is x = - 2. Substituting 2x AX = 3, we get: - 4 + 2A = 3, a = 7 / 2

If the integer solution of the inequality system 2x + 32th (x-3) is the root of the equation 2X-4 = ax about X, find the value of A

1、2x+3-3
3. Because the integer solution of X is: - 2, take the value into 2X-4 = ax,
a＝－8/-2=4