The following equation is solved by means of method: 1. X ^ + X-12 = 0 2. X ^ + 4x + 8 = 2x + 11 3. X (x-4) = 2-8x 3. X ^ + 2x = 0

The following equation is solved by means of method: 1. X ^ + X-12 = 0 2. X ^ + 4x + 8 = 2x + 11 3. X (x-4) = 2-8x 3. X ^ + 2x = 0

X ^ 2 + X-12 = (x + 4) (x-3) = 0, x = 3, x = - 4x ^ 2 + 4x + 8 = 2x + 11 x ^ 2 + 2x-3 = (x + 3) (x-1) = 0, x = 1, x = - 3x (x-4) = 2-8x x ^ 2 + 4x-2 = 0 x = 2 + 2 + radical 6, x = 2-radical 6x ^ 2 + 2x = 0 x (x + 2) = 0 x = 0, x = - 2

It is known that equation 3 [X-2 (x-3 part a)] = 4x and x-3 + a - (1-2x) = 1 have the same solution. Find the value of a and find the solution of the equation

3 [X-2 (x-3 / a)] = 4x --- 1)
3 / x + a - (1-2x) = 1-2)
Bivariate first order equation:
X = 2A / 7 -- 1)
A=2-(7X)/3 ---2)
Substituting 1) into 2) gives:
A＝6/5.

If the solution of equation 4x-3 = 3m-2x x = - two thirds, then M = ()

The solution of equation 4x-3 = 3m-2x x = - 2 / 3
So 4 × (- 2 / 3) - 3 = 3m-2 × (- 2 / 3)
-8/3-3=3m+4/3
3m=-8/3-4/3-3
3m=-4-3
3m=-7
m=-7/3

It is known that the equation (2x-a in 3) - (x-a in 2) = X-1 and equation 3 (X-2) = 4x-6 have the same solution, so find the value of A

By solving equation 2, x = 0
If x = 0 is introduced into equation 1, a = - 6 is obtained

First simplify, then evaluate (3x / X-2 + X / x + 2) / (x / x? - 4), where x = - 3

[3x/(x-2)+x/(x+2)]/[x/(x²-4)]={[3x(x+2)+x(x-2)]/(x²-4)}/[x/(x²-4)]=[(4x²+4x)/(x²-4)]/[x/(x²-4)]=[4x(x+1)/(x²-4)]/[x/(x²-4)]=4x+4=4×(-3)+4=-8

First simplify, then evaluate (x + 2) (X-2) + (X-2) 2 + (x-4) (x-1), where x? - 3x = 1

(x+2)(x-2)+(x-2)²+(x-4)(x-1)
=x²-4+x²-4x+4+x²-5x+4
=3x²-9x+4
=3(x²-3x)+4
=3*1+4
=7

First simplify and then evaluate {(x-2x) - x + 2 / X} △ x-4 2x

Within the brackets, the denominator is x square-4, the numerator is 3x square + 6x, and the other is xsquare-2x
Simplification: {x 2 - 2 x 2 of 4 + 8 x} △ x 2 - 2 x 4
Result = x + 4

First simplify and then evaluate (3x-2x 2) (3-2x-x 2) / (x 2 + x) (2x? - 5x + 3), where x = [2] how to write, it is not easy to simplify at all!

(3x-2x²)(3-2x-x²)/(x²+x)(2x²-5x+3)=x(3-2x)(3+x)(1-x)/x(x+1)(2x-3)(x-1)
=(x+3)/(x+1)=5/3

2x2 - [x2-2 (x2-3x-1) - 3 (x2-1-2x)] where: x = 1 2．

Original formula = 2x2 - (x2-2x2 + 6x + 2-3x2 + 3 + 6x)
=2x2-（-4x2+12x+5）
=6x2-12x-5
∵x=1
2，
The original formula is 1 × 6
4-12×1
2-5=-19
2．

Simplify and then evaluate: (- 2 divided by 3x ^ 3 y? 2) divided by (- 4 divided by 3x? Y?)

(- 2 divided by 3x ^ 3 y? 2) divided by (- 4 divided by 3x??)
=[(- 2) / (3x ^ 3 y 2)] 2 × (- 4 divided by 3x? Y?) / (- 4)
=1/(3x⁴y²)