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\(\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{4-x^2}\right)-\dfrac{2-x}{2+x}:\dfrac{-\left(x-1\right)}{2x-x^2}\)

\(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}\right)+\dfrac{x-2}{x+2}:\dfrac{-\left(x-1\right)}{-x\left(x-2\right)}\)

\(=\dfrac{-\left(x+2\right)^2-4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\cdot\dfrac{x\left(x-2\right)}{x-1}\)

\(=\dfrac{-5x^2-4x-4}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{\left(-5x^2-4x-4\right)\left(x-1\right)+x\left(x-2\right)^3}{\left(x-2\right)\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{-5x^3+5x^2-4x^2+4x-4x+4+x\left(x^3-6x^2+12x-8\right)}{\left(x-2\right)\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{-5x^3+x^2+4+x^4-6x^3+12x^2-8x}{\left(x-2\right)\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{x^4-11x^3+13x^2-8x+4}{\left(x-2\right)\left(x+2\right)\left(x-1\right)}\)

7 tháng 12 2019

d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)

d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)

\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

\(=x^2-2x+1\)

\(=\left(x-1\right)^2\)

23 tháng 9 2021

sao làm có 1 ý vậy bạn ơi

bucqua

a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)

b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)

\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)

\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)

c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)

d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)

\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)

1 tháng 4 2020

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18 tháng 9 2020

tớ chịu.

hi hi.

2 tháng 1 2023

a.(x+10) /(4*x)-8* 4 -(2*x)/x+2

-(127*x-10)/(4*x)

(5/2-127*x/4)/x

2 tháng 1 2023

Câu a