Đáp án: N=x+y-xy
Giải thích các bước giải:
$\begin{array}{l}
N = \frac{{{x^2}}}{{\left( {x + y} \right)\left( {1 - y} \right)}} - \frac{{{y^2}}}{{\left( {x + y} \right)\left( {1 + x} \right)}} - \frac{{{x^2}{y^2}}}{{\left( {1 + x} \right)\left( {1 - y} \right)}}\\
= \frac{{{x^2}\left( {1 + x} \right) - {y^2}\left( {1 - y} \right) - {x^2}{y^2}\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{{x^2} + {x^3} - {y^2} + {y^3} - {x^2}{y^2}\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{{x^2} - {y^2} + \left( {{x^3} + {y^3}} \right) - {x^2}{y^2}\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{\left( {x + y} \right)\left( {x - y} \right) + \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - {x^2}{y^2}\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{\left( {x + y} \right)\left( {x - y + {x^2} - xy + {y^2} - {x^2}{y^2}} \right)}}{{\left( {x + y} \right)\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{x - xy + {y^2} - y + {x^2} - {x^2}{y^2}}}{{\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{x\left( {1 - y} \right) - y\left( {1 - y} \right) + {x^2}\left( {1 - y} \right)\left( {1 + y} \right)}}{{\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{\left( {1 - y} \right)\left( {x - y + {x^2} + {x^2}y} \right)}}{{\left( {1 - y} \right)\left( {1 + x} \right)}}\\
= \frac{{x\left( {x + 1} \right) + y\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {1 + x} \right)}}\\
= \frac{{\left( {x + 1} \right)\left( {x + y\,x - y} \right)}}{{1 + x}}\\
= x - y + xy
\end{array}$
