Solving the fractional equation (3x-6) / (x ^ 2 + X-6) = (4x + 4) / (x ^ 2-3x-4)

Solving the fractional equation (3x-6) / (x ^ 2 + X-6) = (4x + 4) / (x ^ 2-3x-4)

(3x-6)/(x^2+x-6)=(4x+4)/(x^2-3x-4)
3(x-2)/[(x+3)(x-2)]=4(x+1)/[(x-4)(x+1)]
3/(x+3）=4/(x-4)
4(x+3)=3（x-4）
4x+12=3x-12
4x-3x=-12-12
x=-24
It is proved that x = - 24 is the solution of the original equation

It is known that the fraction (3x ^ 4 + 2x ^ 2y-16) / (y + 4) is equal to the value of x ^ 3, and Y ^ 2 + 6x ^ 2 + 4x-96 = 0,5xy-96 = - 2Y, y-2x Finally, if y-2x is not equal to 0, find the value of X

What's the question you're going to ask? How much is x and y?
Given that y ^ 2 + 6x ^ 2 + 4x-96 = 0,5xy-96 = - 2Y
y^2+6x^2+4x=5xy+2y
Displaced
y^2-5xy+6x^2=2y-4x
(y-2x)(y-3x)=2(y-2x)
y-3x=2
y=2+3x
It is also known that (3x ^ 4 + 2x ^ 2y-16) / (y + 4) = x ^ 3
Move (y + 4) to the right
3x^4+2x^2y-16=x^3y+4x^3
Displaced
3x^4+2x^2y-x^3y-4x^3=16
By substituting y = 2 + 3x into the above formula
3x^4+2x^2(2+3x)-x^3(2+3x)-4x^3=16
3x^4+4x^2+6x^3-2x^3-3x^4-4x^3=16
4x^2=16
x^2=4
X = 2 or x = - 2
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If the value of 2x + 1 in fraction (4x-3) is zero, what about the value of 4x + 2 in fraction (2x ^ 2-3x-2)

2x+1=0
2 (2x + 1) / (X-2) (2x + 1) meaningless

(2 X-6) / (x + 3) * (3-x) fractions (x 2 + X-6) (2x-6) / (x + 3) * (3-x) cent (x 2 + X-6) (x 2 - 4 x + 4) + X + 2 (x 2 - 2x) - (1 of x) + 1 [(x 2 - 2x) parts (x + 2) - (x? - 4x + 4) parts (x-1)] / (x? + 4x) parts (x? - 16)

1. 2 (x-3) / (X-2) 2 × 1 / (x + 3) × [- (x + 3) (X-2) / (x-3)] = - 2 / (X-2) 2, (X-2) 2 / (x + 2) (X-2) + X (X-2) / (x + 2) - 1 / x + 1 = (X-2) / (x + 2) + X (X-2) / (x + 2) + (x-1) / X this check 3

Find the simplest common denominator of fractions 4x-2x ^ 2 and x ^ 2-4

1/(4x-2x²)=1/[2x(2-x)]=-1/[2x(x-2)]
1/(x²-4)=1/[(x+2)(x-2)]
-The simplest common denominator of 1 / [2x (X-2)] and 1 / [(x + 2) (X-2)] is 2x (x + 2) (X-2)

Addition and subtraction of fractions: (1 / x-y-x ^ 2 + y ^ 2) - (1 / 1-x ^ 2-2xy-y ^ 2) Note: / "refers to division sign

1/(x-y-x^2+y^2)-1/(1-x^2-2xy-y^2)
=1/[(x-y)(1-x-y)]-1/[1-(x+y)^2]
=1/[(x-y)(1-x-y)]-1/[(1-x-y)(1+x+y)]
=[1+x+y-(x-y)]/[(x-y)(1-x-y)(1+x+y)]
=(1+2y)/[(x-y)(1-x-y)(1+x+y)]

The addition and subtraction of fractions x-2-x-4-x ^ 2 / x ^ 2-4x + 4

=x\(2-x)-(2+x)(2-x)\(2-x)^2 =x\(2-x)-(2+x)\(2-x) =(x-2-x)\(2-x) =-2\(2-x) =2\(x-2)

When x = when, the value of fraction 2x-1 / 4x is 1 / 2 greater than that of fraction X-2 / x + 1

4x/(2x-1)-(x+1)/(x-2)=1
Double (2x-1) (X-2)
4x²-8x-2x²-x+1=2x²-5x+2
4x=-1
x=-1/4

When x = the value of fraction 4x / 2x-1 is equal to that of fraction 2x-1 / X-2

4x/2x-1=2x-1/x-2
（2x-1）*（2x-1）=4x*（x-2）
x=-1/4

If the value of fraction 1 / x ^ 2 + 2x + 7 is 1 / 8, then the value of fraction 1 / 2x ^ 2 + 4x-8 is_____

1/x^2+2x+7=1/8
x^2+2x+7=8
x²+2x=1
∴1/(2x²+4x-8)
=1/[2(x²+2x)-8]
=1/(2-8)
=-1/6