Do not change the values of the following fractions so that the coefficients of the highest phase in the numerator and denominator are positive, and the polynomials in the numerator and denominator are arranged in descending power of X （1）2x+1-x²/-3-2x （2）-（x²y-2xy/4-x²）

Do not change the values of the following fractions so that the coefficients of the highest phase in the numerator and denominator are positive, and the polynomials in the numerator and denominator are arranged in descending power of X （1）2x+1-x²/-3-2x （2）-（x²y-2xy/4-x²）

(1)x²/3+2x-2x+1
(2)x²y-x²-2xy/4

Do not change the values of the following fractions so that the coefficients of the highest order term in the numerator and denominator are positive, and the polynomials in the numerator and denominator are arranged in descending power of X 2X + 1 - x 2 / - 3 - 2x; - x? - 3x + 1 / 2 - x? And explain what the title means

The title is to allow you to make identical transformations of fractions according to their basic properties
And the numerator denominator is arranged according to the descending power of X, that is, the order is from high to low

Without changing the values of fractions, the coefficients of the highest order terms of the numerator and denominator of the following fractions are converted into positive numbers, and the numerator and denominator are arranged in descending power 3a-a/1-a x-x²/1+2x² -1+2x-x²/-2x+1 -Thank you very much

The first question seems to have been copied wrong. I will give you the last three answers (2) (- x? + x) / (1 + 2x?) = - [(x? - x) / (2x? + 1)]. Put the negative sign before the fraction (3) (- x? + 2x-1) / (- 2x + 1) = (x? - 2x + 1) / (2x-1) (4) (- a? + A-1)

Without changing the values of fractions, the coefficients of the highest order terms of the numerator and denominator of the following fractions are converted into positive numbers, and the numerator and denominator are arranged in descending power X -X²/1+2X²

X -X²/1+2X²
=-(x²-x)/(2x²+1)

The polynomials in the numerator and denominator are arranged according to the power reduction of X, and then the value of the fraction is not changed. The coefficient of the highest order term in the numerator and denominator is positive （1）1-x/1-2x-x^2 (2) - 3y-7y^2/5-7y+y^2

（1）1-x/1-2x-x^2
=(x-1)/(x²+2x-1)
(2) - 3y-7y^2/5-7y+y^2
=-(7y²+36y)/(y²-7y+5)

If the value of the fraction 2X-4 x-4 is zero, find the value of X Please write the process clearly

(x²－4)/(2x－4)＝0
∴x²－4＝0
x²＝4
x=±2
∵2x－4≠0
∴x=-2

If the fraction x? 2 + 2X-4 is x? - 2x + 4 = 1, then the value of X is

x²-2x+4=x²+2x-4
4x=8
X=2

If the value of fraction 2x 2 - 5x + 2 / 2 x 2 - 4 is 0, find the value of X

That is (x + 2) (X-2) / (X-2) (2x-1) = 0
(x+2)/(2x-1)=0
So x + 2 = 0
x=-2

Known fraction x2 − 4 If the value of 2x2 − 5x + 2 is 0, find the value of X

According to the meaning of the title: x2-4 = 0, 2x2-5x + 2 ≠ 0,
The solution is: x = 2 (omitted) or x = - 2,
Then the value of X is - 2

The value of X that makes the value of fraction (2x? - 5x-12) / x? 12 to zero is

The value of (2x-5x-12) / X-12 is zero
So 2x? 5x-12 = 0 and X? - X-12 ≠ 0
By solving 2x 2-5x-12 = 0
(2x + 3) (x-4) = 0, x = - 3 / 2 or x = 4
Solving x 2 - X-12 ≠ 0
Then (x-4) (x + 3) ≠ 0 leads to X ≠ - 3 or X ≠ 4
In conclusion, x = - 3 / 2