The square of X + 4x + the square of 1-x divided by the square of (x-1) times the square of (x-1) + 3x + 2

The square of X + 4x + the square of 1-x divided by the square of (x-1) times the square of (x-1) + 3x + 2

(1-x)^2/(x^2+4x+4)/(x-1)^2*(x^2+3x+2)/(x-1)
=(x+1)/(x+2)(x-1)
=(x+1)/(x^2+x-2)

Factorization of (x squared - 4x + 4) (3-x) divided by (square of X - 3x + 6)~~ Factorization of (x squared-4x + 4) (3-x) divided by (square of x-3x + b)~~

(x^2-4x+4)(3-x)/x^2-3x+6=(x-2)^2(3-x)/
X ^ 2-3x + 6 cannot be factorized in real numbers
Classmate, please see if your title is wrong

The square of X - 9 times the square of X - 4x + 4 divided by the square of X + 3 times the square of X - 4 divided by the square of X= Come on, anybody tell me You can count

The square of X - 9 times the square of X - 4x + 4 divided by the square of X + 3 times the square of X - 4 divided by the square of X
=（x-2）²/(x+3)(x-3)÷ （x+2）(x-2)/x(x+3)÷x²
=（x-2）（x+2）/x(x-3)÷x²
=（x²-4）/x³（x-3）

If the fraction 3x * x-x-2 parts 4x-9 = 3x + 2 parts a-x-1 Part B, then the values of a and B are

Right = [a (x-1) - B (3x + 2)] / (x-1) (3x + 2)
=[(a-3b)x+(-a-2b)]/(3x²-x-2)
=(4x-9)/(3x²-x-2)
So (a-3b) x + (- a-2b) = 4x-9
So a-3b = 4
-a-2b=-9
So a = 7, B = 1

When x is the value, the value of fraction x ^ 2-3x + 2 / x ^ 2-4x + 5 is 1

Then there is 3x + 2x-2
x^2-3x+2=x^2-4x+5
-3x+2=-4x+5
4x-3x=5-2
So: x = 3

Solving fractional equation 4 / (1-x ^ 2) = (3x / 1-x) + 3 2 / (X-2) - 4x / (x ^ 2-4) = 0

The first formula, x = 1 / 3
The second equation, there is no solution
4/(1-x^2)=(3x/1-x )+3
4/(1+X)(1-X)=(3X+3-3X)/(1-X)
From the denominator is not equal to 0, we can know that x ≠± 1
X = 1 / 3
2/(x-2)-4x/(x^2-4)=0
2/(x-2)=4x/(x^2-4)
2/(x-2)=4x/（x-2）（x+2）
The solution is x = 2
From the denominator is not equal to 0, we can know that x ≠ ± 2
So there is no solution

The solution of fractional equation 1 / (3x - 6) = 3 / (4x - 8) is obtained

1/（3x-6)=3/（4x-8）
That is, 9x-18 = 4x-8
5x=10
X=2

Given (3x-2) 2 + | 2x-y-3 | = 0, find the value of 5 (2x-y) - 2 (6x-2y + 2) + (4x-3y-2)

2x-y-3 = 0, x = 2 / 3; 2x-y-3 = 0, y = 2 / 3; 2x-y-3 = 0, y = 2x-3 = 4 / 3-3 = - 5 / 35 (2x-y-y) - 2 (6x-2y + 2) + (4x-3y-2 / 1) = 10x-5y-12x + 4y-4 + 4x-3y-1 / 2 = 2x-4y-4-4 + 4x-3y-1 / 2 = 2x-4y-4-1 / 2 = 4 / 3 + 20 / 3-4-4-4-1 / 2 = 24 / 3-4-4-1 / 2 = 24 / 3-4-4-1-1-1-1 / 2 = 24 / 3-3-4-4-1-/ 2 = 8-4-1 / 2 = 4-1 / 2 = 7 / 2

Known: 3x + 2Y = 13, 2x-3y = 0. Find the value of x ^ 2 + X-2 / x ^ 2 + 4x + 4 divided by x ^ 2 + 2x-3 / x ^ 2-4 multiplied by x ^ 2 + 6x + 9 / x ^ 2-xy-2x + 2Y Please help me to figure out the urgent ah

∵3x+2y=13,2x-3y=0 ∴x=3,y=2 [(x^2+x-2)/(x^2+4x+4)]÷[(x^2+2x-3)/(x^2-4)]×[(x^2+6x+9)/(x^2-xy-2x+2y)] =[(x+2)(x-1)/(x+2)(x+2)]×[(x+2)(x-2)/(x+3)(x-1)]×[(x+3)(x+3)/(x-y)(x-2)] =(x+3)/(x-y) =(3+3)/(3-2) =6

Given (3x-2) 2 + | 2x-y-3 | = 0, find the value of 5 (2x-y) - 2 (6x-2y + 2) + (4x-3y-half)

Absolute values and complete squares are nonnegative
So 3x-2 = 0
x=2/3
2x-y-3=0
4/3-y-3=0
y=-5/3
5 (2x-y) - 2 (6x-2y + 2) + (4x-3y-half)
=10x-5y-12x+4y-4+4x-3y-1/2
=2x-4y-4.5
=2x-y-3y-4.5
=4-3*(-5/3)-4.5
=4+5-4.5
=9-4.5
=4.5
Note here: 2x-y-3 = 0
2x-y=3