Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{{x^2} + 3xy}}{{{x^2} - 9{y^2}}} + \frac{{2{x^2} - 5xy - 3{y^2}}}{{6xy - {x^2} - 9{y^2}}}\\
= \frac{{x\left( {x + 3y} \right)}}{{\left( {x - 3y} \right)\left( {x + 3y} \right)}} + \frac{{2{x^2} - 5xy - 3{y^2}}}{{ - \left( {{x^2} - 6xy + 9{y^2}} \right)}}\\
= \frac{x}{{x - 3y}} + \frac{{2{x^2} - 5xy - 3{y^2}}}{{ - {{\left( {x - 3y} \right)}^2}}}\\
= \frac{{x\left( {x - 3y} \right)}}{{{{\left( {x - 3y} \right)}^2}}} - \frac{{2{x^2} - 5xy - 3{y^2}}}{{{{\left( {x - 3y} \right)}^2}}}\\
= \frac{{{x^2} - 3xy - \left( {2{x^2} - 5xy - 3{y^2}} \right)}}{{{{\left( {x - 3y} \right)}^2}}}\\
= \frac{{ - {x^2} + 2xy + 3{y^2}}}{{{{\left( {x - 3y} \right)}^2}}} = \frac{{\left( { - x + 3y} \right)\left( {x + y} \right)}}{{{{\left( {x - 3y} \right)}^2}}} = \frac{{x + y}}{{3y - x}}\\
\frac{{{x^2} + xz + xy + yz}}{{3yz - {x^2} - xz + 3xy}} = \frac{{x\left( {x + z} \right) + y\left( {x + z} \right)}}{{3y\left( {x + z} \right) - x\left( {x + z} \right)}} = \frac{{\left( {x + z} \right)\left( {x + y} \right)}}{{\left( {x + z} \right)\left( {3y - x} \right)}} = \frac{{x + y}}{{3y - x}}\\
\Rightarrow \frac{{{x^2} + 3xy}}{{{x^2} - 9{y^2}}} + \frac{{2{x^2} - 5xy - 3{y^2}}}{{6xy - {x^2} - 9{y^2}}} = \frac{{{x^2} + xz + xy + yz}}{{3yz - {x^2} - xz + 3xy}}
\end{array}\)
