Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{5{x^2} + 2}}{{2x - 5}}\\
\dfrac{{8x + 7}}{{5 - 2x}} = \dfrac{{8x + 7}}{{ - \left( {2x - 5} \right)}} = \dfrac{{ - 8x - 7}}{{2x - 5}}\\
c,\\
\dfrac{5}{{{x^2} + 10x + 25}} = \dfrac{5}{{{x^2} + 2.x.5 + {5^2}}} = \dfrac{5}{{{{\left( {x + 5} \right)}^2}}} = \dfrac{{10}}{{2.{{\left( {x + 5} \right)}^2}}}\\
\dfrac{{x - 5}}{{2x + 10}} = \dfrac{{x - 5}}{{2.\left( {x + 5} \right)}} = \dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{2.{{\left( {x + 5} \right)}^2}}} = \dfrac{{{x^2} - {5^2}}}{{2.{{\left( {x + 5} \right)}^2}}} = \dfrac{{{x^2} - 25}}{{2.{{\left( {x + 5} \right)}^2}}}\\
d,\\
\dfrac{{x + 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} + 3x + 2}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{{x + 3}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \dfrac{{\left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}} = \dfrac{{{x^2} + x - 6}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}
\end{array}\)
