Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{5{x^2} + 10xy + 5{y^2}}}{{3{x^2} + 3{y^2}}} = \frac{{5.\left( {{x^2} + 2xy + {y^2}} \right)}}{{3.\left( {{x^2} + {y^2}} \right)}} = \frac{{5{{\left( {x + y} \right)}^2}}}{{3.\left( {{x^2} + {y^2}} \right)}}\\
\frac{{4{x^4} - 20{x^3} + 13{x^2} + 30x + 9}}{{{{\left( {4{x^2} - 1} \right)}^2}}}\\
= \frac{{\left( {4{x^4} + 2{x^3}} \right) - \left( {22{x^3} + 11{x^2}} \right) + \left( {24{x^2} + 12x} \right) + \left( {18x + 9} \right)}}{{{{\left[ {\left( {2x - 1} \right)\left( {2x + 1} \right)} \right]}^2}}}\\
= \frac{{2{x^3}\left( {2x + 1} \right) - 11{x^2}\left( {2x + 1} \right) + 12x\left( {2x + 1} \right) + 9\left( {2x + 1} \right)}}{{{{\left( {2x - 1} \right)}^2}.{{\left( {2x + 1} \right)}^2}}}\\
= \frac{{\left( {2x + 1} \right).\left( {2{x^3} - 11{x^2} + 12x + 9} \right)}}{{{{\left( {2x - 1} \right)}^2}.{{\left( {2x + 1} \right)}^2}}}\\
= \frac{{\left( {2x + 1} \right).\left[ {\left( {2{x^3} + {x^2}} \right) - \left( {12{x^2} + 6x} \right) + \left( {18x + 9} \right)} \right]}}{{{{\left( {2x - 1} \right)}^2}.{{\left( {2x + 1} \right)}^2}}}\\
= \frac{{\left( {2x + 1} \right).\left( {2x + 1} \right)\left( {{x^2} - 6x + 9} \right)}}{{{{\left( {2x - 1} \right)}^2}.{{\left( {2x + 1} \right)}^2}}}\\
= \frac{{{{\left( {2x + 1} \right)}^2}{{\left( {x - 3} \right)}^2}}}{{{{\left( {2x - 1} \right)}^2}{{\left( {2x + 1} \right)}^2}}}\\
= {\left( {\frac{{x - 3}}{{2x - 1}}} \right)^2}
\end{array}\)
