Rút gọn biểu thức D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
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20 tháng 11 2023
1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
1 tháng 1 2022
\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
17 tháng 4 2021
Ta có: \(\dfrac{x^2+xy}{x^2+xy+y^2}-\left(\dfrac{x\left(2x^2+xy-y^2\right)}{x^3-y^3}-2+\dfrac{y}{y-x}\right):\dfrac{x-y}{x}-\dfrac{x}{x-y}\)
\(=\dfrac{x^2+xy}{x^2+xy+y^2}-\left(\dfrac{x\left(2x^2+xy-y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{2\left(x^3-y^3\right)-y\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\right):\dfrac{x-y}{x}-\dfrac{x}{x-y}\)
\(=\dfrac{x^2+xy}{x^2+xy+y^2}-\dfrac{2x^3+x^2y-xy^2-2x^3+2y^3-x^2y-xy^2-y^3}{\left(x-y\right)\left(x^2+xy+y^2\right)}:\dfrac{x-y}{x}-\dfrac{x}{x-y}\)
\(=\dfrac{x\left(x+y\right)}{x^2+xy+y^2}-\dfrac{y^3-2xy^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}:\dfrac{x-y}{x}-\dfrac{x}{x-y}\)
\(=\dfrac{x\left(x+y\right)}{x^2+xy+y^2}+\dfrac{y^2\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\cdot\dfrac{x}{x-y}-\dfrac{x}{x-y}\)
\(=\dfrac{x\left(x+y\right)}{x^2+xy+y^2}+\dfrac{xy^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{x}{x-y}\)
\(=\dfrac{x\left(x^2-y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{xy^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{x\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^3-xy^2+xy^2-x^3-x^2y-xy^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{-x^2y-xy^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
25 tháng 2 2021
`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`
`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`
`=(y(y-2x))/3`
`b,(x^2-y^2)/(x^2-y^2+xz-yz)`
`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`
`=(x+y)/(x+y+z)`
`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`
`=(-(x^2-3x+x-3))/((x-1)(x+1))`
`=(-x(x-3)+x-3)/((x-1)(x+1))`
`=((x-3)(1-x))/((x-1)(x+1))`
`=(3-x)/(1+x)`
`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`
`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`
`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`
`=(3x^2-4x+1)/(2x^2+x-3)`
`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`
`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`
`=(3x-1)/(2x+3)`
25 tháng 2 2021
a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)
\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)
\(=\dfrac{y\left(y-2x\right)}{3}\)
\(D=\dfrac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\dfrac{y^2\left(x+y\right)}{x}\\ D=\dfrac{\left(x^2-y^2\right)\left(x+y\right)+y^2\left(x+y\right)}{x}\\ D=\dfrac{\left(x+y\right)\left(x^2-y^2+y^2\right)}{x}=\dfrac{x^2\left(x+y\right)}{x}=x\left(x+y\right)\)