$M+\left( 5{{x}^{2}}-2xy \right)=6{{x}^{2}}-9xy-{{y}^{2}}$

$M=6{{x}^{2}}-9xy-{{y}^{2}}-\left( 5{{x}^{2}}-2xy \right)$

$M={{x}^{2}}-7xy-{{y}^{2}}$

Ta có ${{\left( 2x-5 \right)}^{2018}}\ge 0;\,\,{{\left( 3y+4 \right)}^{2020}}\ge 0$ với mọi $x;y$ nên

${{\left( 2x-5 \right)}^{2018}}+{{\left( 3y+4 \right)}^{2020}}=0$

$\Leftrightarrow \left\{ \begin{align}& 2x-5=0 \\ & 3y+4=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=\frac{5}{2} \\ & y=-\frac{4}{3} \\ \end{align} \right.$

Do đó : $M={{x}^{2}}-7xy-{{y}^{2}}={{\left( \frac{5}{2} \right)}^{2}}-7.\frac{5}{2}.\left( -\frac{4}{3} \right)-{{\left( -\frac{4}{3} \right)}^{2}}=$ học sinh tự tính .