Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{{x^2} + 2x + 4}}{{4{x^3} - 32}} = \dfrac{{{x^2} + 2x + 4}}{{4.\left( {{x^3} - 8} \right)}} = \dfrac{{{x^2} + 2x + 4}}{{4.\left( {{x^3} - {2^3}} \right)}}\\
= \dfrac{{{x^2} + 2x + 4}}{{4.\left( {x - 2} \right)\left( {{x^2} + x.2 + {2^2}} \right)}} = \dfrac{{{x^2} + 2x + 4}}{{4.\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{1}{{4.\left( {x - 2} \right)}}\\
b,\\
\dfrac{{{x^2} + 3xy}}{{{x^2} - 9{y^2}}} = \dfrac{{x.\left( {x + 3y} \right)}}{{{x^2} - {{\left( {3y} \right)}^2}}} = \dfrac{{x\left( {x + 3y} \right)}}{{\left( {x - 3y} \right)\left( {x + 3y} \right)}} = \dfrac{x}{{x - 3y}}\\
c,\\
\dfrac{{{x^2} + 4{y^2} - 4xy - 4}}{{4{x^2} - 4xy + 4x}} = \dfrac{{\left( {{x^2} - 4xy + 4{y^2}} \right) - 4}}{{4x.\left( {x - y + 1} \right)}}\\
= \dfrac{{\left( {{x^2} - 2.x.2y + {{\left( {2y} \right)}^2}} \right) - {2^2}}}{{4x.\left( {x - y + 1} \right)}} = \dfrac{{{{\left( {x - 2y} \right)}^2} - {2^2}}}{{4x.\left( {x - y + 1} \right)}}\\
= \dfrac{{\left( {x - 2y - 2} \right)\left( {x - 2y + 2} \right)}}{{4x.\left( {x - y + 1} \right)}}
\end{array}\)
Em xem lại đề câu c nhé em?
