Đáp án:

f) \(\dfrac{1}{{{{\left( {x + 2} \right)}^2}}} = \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{{13z}}{{63{x^2}{y^3}}} = \dfrac{{13.5{z^3}}}{{63.5{x^2}{y^3}{z^2}}} = \dfrac{{65{z^3}}}{{315{x^2}{y^3}{z^2}}}\\
 - \dfrac{y}{{15x{z^2}}} =  - \dfrac{{3.7.x.{y^3}}}{{315{x^2}{y^3}{z^2}}} =  - \dfrac{{21x.{y^3}}}{{315{x^2}{y^3}{z^2}}}\\
\dfrac{{2x}}{{9{y^2}z}} = \dfrac{{2.5.7.{x^2}.yz}}{{315{x^2}{y^3}{z^2}}} = \dfrac{{70{x^2}yz}}{{315{x^2}{y^3}{z^2}}}\\
b)\dfrac{x}{{x - y}} = \dfrac{{x{{\left( {x - y} \right)}^2}}}{{{{\left( {x - y} \right)}^3}}}\\
\dfrac{y}{{{{\left( {x - y} \right)}^2}}} = \dfrac{{y\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^3}}}\\
\dfrac{1}{{{{\left( {y - x} \right)}^3}}} =  - \dfrac{1}{{{{\left( {x - y} \right)}^3}}}\\
c)\dfrac{1}{{2\left( {x + 2} \right)}} = \dfrac{{x - 2}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{x}{{2\left( {x - 2} \right)}} = \dfrac{{x\left( {x + 2} \right)}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} + 2x}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{3}{{4 - {x^2}}} = \dfrac{{ - 3.2}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}} =  - \dfrac{6}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}\\
d)\dfrac{1}{{x\left( {x - 2} \right)}} =  - \dfrac{{\left( {x + 2} \right)\left( {2x + 1} \right)}}{{x\left( {2 - x} \right)\left( {x + 2} \right)\left( {2x + 1} \right)}}\\
\dfrac{{20}}{{x\left( {2 - x} \right)\left( {2 + x} \right)}} = \dfrac{{20\left( {2x + 1} \right)}}{{x\left( {2 - x} \right)\left( {x + 2} \right)\left( {2x + 1} \right)}}\\
\dfrac{7}{{x\left( {2x + 1} \right)}} = \dfrac{{7\left( {2 - x} \right)\left( {x + 2} \right)}}{{x\left( {2 - x} \right)\left( {x + 2} \right)\left( {2x + 1} \right)}}
\end{array}\)

\(\begin{array}{l}
e)\dfrac{x}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \dfrac{{{x^2}}}{{x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
\dfrac{{x + 1}}{{x\left( {x + 1} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \dfrac{{{x^3} + 1}}{{x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
\dfrac{{x + 2}}{{{x^2} - x + 1}} = \dfrac{{x\left( {x + 1} \right)\left( {x + 2} \right)}}{{x\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
f)\dfrac{1}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}} = \dfrac{{{x^2} + 3x + 2}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}}\\
\dfrac{1}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}} = \dfrac{{{x^2} + 4x + 4}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}}\\
\dfrac{1}{{{{\left( {x + 2} \right)}^2}}} = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}} = \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x + 2} \right)}^2}{{\left( {x + 1} \right)}^2}}}
\end{array}\)