Đáp án:
4) \({\left( {x + y} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\dfrac{{5\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{3\left( {x + 1} \right)}}{{20\left( {1 - x} \right)}}\\
= \dfrac{{5\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{ - 3\left( {x + 1} \right)}}{{20\left( {x - 1} \right)}}\\
= - \dfrac{3}{{4\left( {x + 1} \right)}}\\
3)\dfrac{{\left( {x + 4} \right)}}{{5x\left( {x - 5} \right)}}.\dfrac{{x - 5}}{{{{\left( {x + 4} \right)}^2}}} = \dfrac{1}{{5x\left( {x + 4} \right)}}\\
2)\dfrac{x}{{x + 1}} + \dfrac{{x - 1}}{x} + \dfrac{x}{{x + 1}}\\
= \dfrac{{2x}}{{x + 1}} + \dfrac{{x - 1}}{x}\\
= \dfrac{{2{x^2} + {x^2} - 1}}{{x\left( {x + 1} \right)}}\\
= \dfrac{{3{x^2} - 1}}{{x\left( {x + 1} \right)}}\\
4)\dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{{x^2} + {y^2}}}:\dfrac{{\left( {{x^2} - 2xy + {y^2}} \right)}}{{\left( {{x^2} - {y^2}} \right)\left( {{x^2} + {y^2}} \right)}}\\
= \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{{x^2} + {y^2}}}:\dfrac{{{{\left( {x - y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\\
= \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{{x^2} + {y^2}}}.\dfrac{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}{{{{\left( {x - y} \right)}^2}}}\\
= {\left( {x + y} \right)^2}
\end{array}\)
