Đáp án:
k) \(\dfrac{{{x^3} + {y^3}}}{{x\left( {{x^3} - {y^3}} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x\left( {4 - x} \right)}} = - \dfrac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x\left( {x - 4} \right)}} = - \dfrac{{x + 4}}{x}\\
b)\dfrac{{x\left( {x + 1} \right) + 3\left( {x + 1} \right)}}{{2\left( {x + 3} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right)}}{{2\left( {x + 3} \right)}} = \dfrac{{x + 1}}{2}\\
c)\dfrac{{15x{{\left( {x + y} \right)}^3}}}{{5y{{\left( {x + y} \right)}^2}}} = \dfrac{{3x\left( {x + y} \right)}}{y}\\
d)\dfrac{{5\left( {x - y} \right) - 3\left( {y - x} \right)}}{{10\left( {x - y} \right)}} = \dfrac{{5\left( {x - y} \right) + 3\left( {x - y} \right)}}{{10\left( {x - y} \right)}}\\
= \dfrac{{8\left( {x - y} \right)}}{{10\left( {x - y} \right)}} = \dfrac{4}{5}\\
e)\dfrac{{2\left( {x + y} \right) + 5\left( {x + y} \right)}}{{2\left( {x + y} \right) - 5\left( {x + y} \right)}} = \dfrac{{7\left( {x + y} \right)}}{{ - 3\left( {x + y} \right)}} = - \dfrac{7}{3}\\
f)\dfrac{{x\left( {x - y} \right)}}{{3y\left( {x - y} \right)}} = \dfrac{x}{{3y}}\\
g)\dfrac{{2a\left( {{x^2} - 2x + 1} \right)}}{{5b\left( {1 - {x^2}} \right)}} = - \dfrac{{2a{{\left( {x - 1} \right)}^2}}}{{5b\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= - \dfrac{{2a\left( {x - 1} \right)}}{{5b\left( {x + 1} \right)}}\\
h)\dfrac{{4x\left( {x - y} \right)}}{{5{x^2}\left( {x - y} \right)}} = \dfrac{4}{{5x}}\\
i)\dfrac{{\left( {x + y - z} \right)\left( {x + y + z} \right)}}{{x + y + z}} = x + y - z\\
k)\dfrac{{{{\left( {{x^3} + {y^3}} \right)}^2}}}{{x\left( {{x^6} - {y^6}} \right)}} = \dfrac{{{{\left( {{x^3} + {y^3}} \right)}^2}}}{{x\left( {{x^3} + {y^3}} \right)\left( {{x^3} - {y^3}} \right)}}\\
= \dfrac{{{x^3} + {y^3}}}{{x\left( {{x^3} - {y^3}} \right)}}
\end{array}\)
