XIN CHỈ GIÚP MÌNH Ạ $ extcircled{a)}\frac{9(x^2-2x+1)}{5y^2}.\frac{10(y+1)-10}{6(x^2-2x)+6}$ $b)\frac{5x^3-40}{x^2-9}.\frac{3x-x^2}{x^2+2x+4}$,$ extcircled c)\frac{2(3-x)}{x^2+2}.\frac{3x+1}{x-1}.\frac{x-3}{2x^2+4}$ $ extcircled{d)}\frac{8x^3}{21y^2}:(\frac{-4x^2}{7y^3})$,$e)\frac1{x^2+y^2}(\frac{x^2+y^2}{x^2y^2}-x^2-y^2)-\frac1{x^3}:\frac{y^2}x$ $f)(\frac xy-\frac yx):[\frac{(x+y)^2}{xy}-2-\frac{2y}x]$,$ extcircled{g)}(\frac x{x+y}+\frac y{x-y}):[\frac{2(x+y)}{x-y}-\frac{4xy}{x^2-y^2}]$,$h)\frac y{x-y}-\frac{x^3-xy^2}{x^2+y^2}.[\frac x{(x-y)^2}-\frac y{x^2-y^2}]$,$i)\frac x{3-x}+\frac{x^2+3x}{2x+3}.(\frac{x+3}{x^2-3x}-\frac x{x^2-9})$,$k)(\frac x{x^2-36}-\frac{x-6}{x^2+6x}):\frac{2x-6}{x^2+6x}+\frac x{6-x}$

Question

XIN CHỈ GIÚP MÌNH Ạ $	extcircled{a)}\frac{9(x^2-2x+1)}{5y^2}.\frac{10(y+1)-10}{6(x^2-2x)+6}$ $b)\frac{5x^3-40}{x^2-9}.\frac{3x-x^2}{x^2+2x+4}$,$	extcircled c)\frac{2(3-x)}{x^2+2}.\frac{3x+1}{x-1}.\frac{x-3}{2x^2+4}$ $	extcircled{d)}\frac{8x^3}{21y^2}:(\frac{-4x^2}{7y^3})$,$e)\frac1{x^2+y^2}(\frac{x^2+y^2}{x^2y^2}-x^2-y^2)-\frac1{x^3}:\frac{y^2}x$ $f)(\frac xy-\frac yx):[\frac{(x+y)^2}{xy}-2-\frac{2y}x]$,$	extcircled{g)}(\frac x{x+y}+\frac y{x-y}):[\frac{2(x+y)}{x-y}-\frac{4xy}{x^2-y^2}]$,$h)\frac y{x-y}-\frac{x^3-xy^2}{x^2+y^2}.[\frac x{(x-y)^2}-\frac y{x^2-y^2}]$,$i)\frac x{3-x}+\frac{x^2+3x}{2x+3}.(\frac{x+3}{x^2-3x}-\frac x{x^2-9})$,$k)(\frac x{x^2-36}-\frac{x-6}{x^2+6x}):\frac{2x-6}{x^2+6x}+\frac x{6-x}$

Accepted Answer

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
a) \ \frac{9\left( x^{2} -2x+1ight)}{5y^{2}} .\frac{10( y+1) -10}{6\left( x^{2} -2xight) +6}\
=\frac{9( x-1)^{2}}{5y^{2}} .\frac{10.( y+1-1)}{6\left( x^{2} -2x+1ight)}\
=\frac{9( x-1)^{2}}{5y^{2}} .\frac{10y}{6( x-1)^{2}}\
=\frac{3}{y}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
c,\frac{2( 3-x)}{x^{2} +2} .\frac{3x+1}{x-1} :\frac{x-3}{2x^{2} +4}\
=\frac{2( 3-x)}{x^{2} +2} .\frac{3x+1}{x-1} .\frac{2\left( x^{2} +2ight)}{x-3}\
=\frac{-4( 3x+1)}{x-1}
\end{array}$
$\displaystyle d,\frac{8x^{3}}{21y^{2}} :\left(\frac{-4x^{2}}{7y^{3}}ight) =\frac{8x^{3}}{21y^{2}} .\frac{7y^{3}}{-4x^{2}} =\frac{-2}{3} xy$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
g,\left(\frac{x}{x+y} +\frac{y}{x-y}ight) :\left[\frac{2( x+y)}{x-y} -\frac{4xy}{x^{2} -y^{2}}ight]\
=\frac{x( x-y) +y( x+y)}{( x+y)( x-y)} :\left[\frac{2( x+y)}{x-y} -\frac{4xy}{( x-y)( x+y)}ight]\
=\frac{x^{2} +y^{2}}{( x-y)( x+y)} :\frac{2( x+y)^{2} -4xy}{( x-y)( x+y)}\
=\frac{x^{2} +y^{2}}{( x-y)( x+y)} .\frac{( x-y)( x+y)}{2\left( x^{2} +y^{2}ight)}\
=\frac{1}{2}
\end{array}$