Ta có:

+) \(\frac{{{x^2} + 3xy}}{{{x^2} - 9{y^2}}} + \frac{{2{x^2} - 5xy - 3{y^2}}}{{6xy - {x^2} - 9{y^2}}}\)

\( = \frac{{x\left( {x + 3y} \right)}}{{\left( {x - 3y} \right)\left( {x + 3y} \right)}} - \frac{{2{x^2} - 6xy + xy - 3{y^2}}}{{{{\left( {x - 3y} \right)}^2}}}\)

\( = \frac{x}{{x - 3y}} - \frac{{2x\left( {x - 3y} \right) + y\left( {x - 3y} \right)}}{{{{\left( {x - 3y} \right)}^2}}}\)

\( = \frac{x}{{x - 3y}} - \frac{{\left( {2x + y} \right)\left( {x - 3y} \right)}}{{{{\left( {x - 3y} \right)}^2}}}\)

\( = \frac{x}{{x - 3y}} - \frac{{2x + y}}{{x - 3y}}\)

\( = \frac{{ - x - y}}{{x - 3y}}\)

\( = \frac{{x + y}}{{3y - x}}\) (1)

+) \(\frac{{{x^2} + xz + xy + yz}}{{3yz - {x^2} - xz + 3xy}}\)

\( = \frac{{x\left( {x + z} \right) + y\left( {x + z} \right)}}{{3y\left( {z + x} \right) - x\left( {x + z} \right)}}\)

\( = \frac{{\left( {x + z} \right)\left( {x + y} \right)}}{{\left( {x + z} \right)\left( {3y - x} \right)}}\)

\( = \frac{{x + y}}{{3y - x}}\) (2)

Từ (1) và (2) suy ra \(\frac{{{x^2} + 3xy}}{{{x^2} - 9{y^2}}} + \frac{{2{x^2} - 5xy - 3{y^2}}}{{6xy - {x^2} - 9{y^2}}} = \frac{{{x^2} + xz + xy + yz}}{{3yz - {x^2} - xz + 3xy}}\)