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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: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
\(A=\left(\frac{4x}{x^2-4}+\frac{2x-4}{x+2}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\=\left(\frac{4x}{x^2-4}+\frac{\left(2x-4\right)\left(x-2\right)}{x^2-4}\right)\frac{x+2}{2x}+\frac{2}{2-x}=\left(\frac{4x}{x^2-4}+\frac{2x^2-4x-4x+8}{x^2-4}\right) \frac{x+2}{2x}+\frac{2}{2-x}\)
\(=\left(\frac{4x+2x^2-8x+8}{x^2-4}\right).\frac{x+2}{2x}+\frac{2}{2-x}\\ =\frac{2x\left(x+2\right)-8\left(x-1\right)}{x^2-4}.\frac{x+2}{2x}+\frac{2}{2-x}\)
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)
\(=\frac{16+x-18}{x\left(x-2\right)}\)
\(=\frac{-2+x}{x\left(x-2\right)}\)
a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)
b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)
c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)
\(\dfrac{x^2-4}{x^2+2x}=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x-2}{x}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}\)
\(\dfrac{x}{x-2}=\dfrac{x^2}{x\left(x-2\right)}\)
a,\(\frac{2x^2+4x}{x+2}\)=\(\frac{2x\left(x+2\right)}{x+2}\)\(=2x\)
b, \(\frac{3x}{2x+4}\)=\(\frac{3x^2-6x}{2\left(x+2\right)\left(x-2\right)}\)
\(\frac{x+3}{x^2+4}\)=\(\frac{2x+6}{2\left(x-2\right)\left(x+2\right)}\)
tick mình nhé!!