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14 tháng 12 2018

\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)

\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)

\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)

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

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

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

b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)

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

6 tháng 10 2021

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

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

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)

12 tháng 9 2017

Đặt \(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+x+xy+y}\right]:\dfrac{x+1}{2x^2+y+2}\)

\(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{\left(x+y\right).\left(x-2y\right)}\right):\dfrac{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}{\left(x+y\right).\left(x+1\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)

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

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

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

Thay \(x=-1,76\)\(y=\dfrac{3}{25}\) vào biểu thức ta được:

\(A=\dfrac{1}{2.\dfrac{3}{25}-\left(-1,76\right)}\)

\(A=\dfrac{1}{2}\)

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

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