\(\dfrac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}=\dfrac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)

\(=\dfrac{29x^2+29}{x^2+1}=\dfrac{29\left(x^2+1\right)}{x^2+1}=29\)

Vậy.....

Ta có: \(\dfrac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}\)

\(=\dfrac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)

\(=\dfrac{29x^2+29}{x^2+1}=29\)

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

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

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

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

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

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

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

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

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

`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)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

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

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))

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

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

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

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)

\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)

\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)

\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)

\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)

\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

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

\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

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

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

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

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

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

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

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

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

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

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

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

\(=1\)