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8 tháng 11 2017

\(\dfrac{x}{x-y}-\dfrac{1}{x-y}-\dfrac{1-y}{y-x}=\dfrac{x}{x-y}-\dfrac{1}{x-y}+\dfrac{y-1}{x-y}=\dfrac{x-1+y-1}{x-y}=\dfrac{x+y-2}{x-y}\)

14 tháng 12 2021

\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)

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

\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)

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

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

4 tháng 12 2017

\(\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}\)

\(=\dfrac{y}{xy\left(y-x\right)}-\dfrac{x}{xy\left(y-x\right)}\)

\(=\dfrac{y-x}{xy\left(y-x\right)}\)

\(=\dfrac{1}{xy}\)

\(\Rightarrow\) Đpcm.

3 tháng 11 2018

A \(\dfrac{3}{x-y}\)

b \(\dfrac{5}{x+y}\)

c \(\dfrac{2x-x^2}{x+1}\)

18 tháng 2 2022

\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)

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

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

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

a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)

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

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

\(=x-1\)

b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)

\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{-1}{y}\)

 

16 tháng 11 2021

a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)