\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)

\(=\frac{1}{x+y}-\frac{3xy}{x^3-y^3}+\frac{x-y}{x^2+xy+y^2}\)

\(=\frac{1.\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\frac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

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

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

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

\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)

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

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

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

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

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

\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)

Ta có:

\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\) \(\left(x\ne y\right)\)

\(=\frac{1}{x-y}-\frac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\frac{x-y}{x^2+xy+y^2}\)

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

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

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

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

a: Sửa đề: \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x-1}\)

\(=\frac{x^2\left(x-1\right)+2x+x+1}{\left(x-1\right)\left(x+1\right)}=\frac{x^3-x^2+3x+1}{\left(x-1\right)\left(x+1\right)}\)

b: \(\frac{2x+y}{2x^2-xy}+\frac{y}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)

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

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

\(=\frac{8x^2-xy+2y^2}{4x^2-y^2}\)

c: \(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)

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

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

a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)

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

\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)

a: \(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\)

\(=\frac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}=\frac{1-2\sqrt{x}+x}{1-\sqrt{x}}\)

\(=\frac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}=1-\sqrt2\)

b: \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{1+\sqrt{xy}}\)

\(=\frac{x-2\sqrt{xy}+y+4\sqrt{xy}}{1+\sqrt{xy}}=\frac{x+2\sqrt{xy}+y}{1+\sqrt{xy}}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{1+\sqrt{xy}}=\frac{\left(\sqrt2+\sqrt3\right)^2}{1+\sqrt{2\cdot3}}=\frac{5+2\sqrt6}{1+\sqrt6}\)

\(=\frac{\left(5+2\sqrt6\right)\left(\sqrt6-1\right)}{5}=\frac{5\sqrt6-5+12-2\sqrt6}{5}=\frac{3\sqrt6+7}{5}\)