Using the addition and subtraction of fractions (2) to calculate (x / x + y + 2Y / x + y) * XY / x + 2Y △ (1 / x + 1 / y) （1）(x/x+y+2y/x+y)*xy/x+2y÷(1/x+1/y) （2） (1/a+1/b）²÷（1/a²-1/b²) （3） (3x²/4y)² * 2y/3x+x²/2y²÷2y²/x （4） (a+b/a-b)² * 2a-2b/3a+3b-a²/a²-b²÷a/b

Using the addition and subtraction of fractions (2) to calculate (x / x + y + 2Y / x + y) * XY / x + 2Y △ (1 / x + 1 / y) （1）(x/x+y+2y/x+y)*xy/x+2y÷(1/x+1/y) （2） (1/a+1/b）²÷（1/a²-1/b²) （3） (3x²/4y)² * 2y/3x+x²/2y²÷2y²/x （4） (a+b/a-b)² * 2a-2b/3a+3b-a²/a²-b²÷a/b

The main way to learn is by oneself. (1) the original formula = (x + 2Y) / (x + y) / (x + 2Y) / (x + 2Y) / (x + y) / x = x / (x + y) / (x + y) / (x + y + y) = (x ^ 2 + 2XY + y ^ 2 (2) original formula = [(a + b) / AB] ^ 2 ^ (b ^ 2-A ^ 2) / A ^ 2B ^ 2 = (a + b) ^ 2 / A ^ 2 ^ 2 * a ^ 2 ^ 2 / [(a + b) (B-A)] = (a + b) / (B-A) (3) primitive = (a + b) / (B-A) (3) is the original primitive primitive (a + b) / (B-A) (3) the original is the original original (a + formula = 9x ^ 4 / 16y ^

(fast) fraction operation: (1) (x + y) · x ^ 2-y ^ 2 parts x ^ 2 + X-Y parts y ^ 2 - (2x ^ 2y-2xy ^ 2) / (x ^ 2-2xy + y ^ 2)

(1) (x + y) · x ^ 2-y ^ 2 parts x ^ 2 + X-Y parts y ^ 2 - (2x ^ 2y-2xy ^ 2) / (x ^ 2-2xy + y ^ 2) = (x + y) * x / (x + y) (X-Y) + Y / (X-Y) - 2XY (X-Y) / (X-Y) = (X-Y) + Y / / (X-Y) - 2XY / (X-Y) = (X-Y + y-2xy) / (X-Y) = (x -...)

How to solve the problem of fraction calculation quickly (x minus 1 of Y + X + 1 of Y) divided by the square of x minus 2XY + the square of Y by 2x= (1) (x minus 1 of Y + X + 1 of Y) divided by the square of x minus 2XY + 2x of Y equals (2) 1 minus (a - 1 - a 1) divided by the square of a - 2A + 1 of a squared - A + 1= (3) (2a's Square b) is multiplied by the minus third power of (A's minus first power and B's negative second power)=

Is there any bracket after "divide"
(x minus 1 of Y + X + 1 of Y) divided by the square of x minus 2XY + the square of Y by 2x
=(x-1/y+x+1/y)÷(x²-2xy+2x/y²)
=2x÷(x²-2xy+2x/y²)
=2x÷[(x²y²-2xy³+2x)/y²]
=2x[y²/(x²y²-2xy³+2x)]
=2xy²/(x²y²-2xy³+2x)
=2y²/(xy²-2y³+2)
If not:
=(x-1/y+x+1/y)÷x²-2xy+2x/y²
=2x÷x²-2xy+2x/y²
=2/x-2xy+2x/y²
=2(1/x-xy+1/y²)
=2(y²-x²y³+x)/xy²

Fraction calculation: x ^ 2-4 / x ^ 2 + 6x + 9 △ x + 2 / x + 3 A / (A-2) - 4A / A ^ 2-4 1 / (X-Y) - 1 / (x + y) * x ^ 2-2xy + y ^ 2 / (y + 1) ^ 2 - (Y-1) ^ 2 x^2-4/x^2+6x+9÷x+2/x+3 a/(a-2)-4a/a^2-4 1/(x-y)-1/(x+y)*x^2-2xy+y^2/(y+1)^2-(y-1)^2

x^2-4/x^2+6x+9÷x+2/x+3=(x²-4)/(x²+6x+9)÷(x+2)/(x+3)=(x+2)(x-2)/(x+3)²×(x+3)/(x+2)=(x-2)/(x+3)a/(a-2)-4a/a^2-4=a/(a-2)-4a/(a+2)(a-2)=a(a+2)/(a+2)(a-2)-4a/(a+2)(a-2)=(a²+2a-4a)/(a...

Addition and subtraction of fractions (x + 3 / x ^ 2-9 + X / 6 + x-x ^ 2) · 3-x / 4x-6

(x + 3) / (x + 3) (x-3) (x-3) (x-3) (x + 3) (x-3) (x + 2)] [(3-x) / (4x-6)] = [1 / (x-3) - X / (x-3) (x + 2)]] [(3-x) / (4x-6)] = {[(x + 2) - x] / (x-3) (x-3) (x + 2)} [(x-3) / (4x- 6)] = [2 / (x-3) (x + 2)] [(x-3) / 2 (2x-3)] = - 1 / (x + 2) (2x-3) = - 1 / (x + 2) (2x-3) = - 1 (1) = - 1 / (x + 2) (2x-3) = - 1 (1) = - 1/ (2x? + X-6)

Triple + 4x + 1 / 2x-1

Because 1 / 1 = 1 2x / 1 = 2x 4x thrice / 1 = 4x triple
So 1 / 1-X-1 / 1 + x-2x / 1 + X + - 4x thrice / 1 + X quadruple
=1-x-1+x-2x+x-4x^3+x^4
=x^4-4x^3-x
=x(x^3-4x^2-1)

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

I'm glad you can answer the question
According to the meaning of the title:
x^2+4
＝－4x
x^2+4x+4=0
(x+2)^2=0

Mixed operation of (9-6x + x ^ 2 / x ^ 2-16) / (x-3 / 4-x) * (x ^ 2 + 4x + 4 / 4-x ^ 2) fractions

(9-6x+x^2)/(x^2-16)/(x-3/4-x)*(x^2+4x+4)/(4-x^2)
=(3-x)^2/(x+4)(x-4)*(4-x)/(x-3)*(x+2)^2/(2+x)(2-x)
=〔(3-x)^2*(4-x)*(x+2)^2〕/〔(x+4)(x-4)*(x-3)*(2+x)(2-x)〕
=〔-(x-3)*(x+2)〕/〔(x+4)*(2-x)〕
=〔(x-3)*(x+2)〕/〔(x+4)*(x-2)〕
=(x^2-x-6)/(x^2+2x-8)

When x = what, the fraction 2x-5 4x + 3 has a value of 1

According to the meaning of the title
(4x+3)/(2x-5)=1
4x+3=2x-5
4x-2x=-5-3
2x=-8
x=-4
When x = - 4, the value of 4x + 3 of fraction 2x-5 is 1

Given 4x + 3 = 0 of X, find the value of (2x-x ^ 2) × (x ^ 2-4x + 4) of fraction (x ^ 3 + x ^ 2) = 0

X = - 3 / 4