<p>Tìm x,y biết -10x^2+24x+2xy-y^2-16=0</p>

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
-10x^{2} +24x+2xy-y^{2} -16=0\\
\Rightarrow -10x^{2} +2xy+24x-y^{2} -16=0\\
\Rightarrow -\left( x^{2} -2xy+y^{2}\right) -9x^{2} +24x-16=0\\
\Rightarrow -( x-y)^{2} -\left[( 3x)^{2} -2.3x.4+16\right] =0\\
\Rightarrow -( x-y)^{2} -( 3x-4)^{2} =0\\
\Rightarrow ( x-y)^{2} +( 3x-4)^{2} =0\\
có:\ ( x-y)^{2} \geqslant 0\\
( 3x-4)^{2} \geqslant 0\\
\Rightarrow ( x-y)^{2} +( 3x-4)^{2} \geqslant 0\\
\Rightarrow ( x-y)^{2} +( 3x-4)^{2} =0\ khi\ \begin{cases}
x-y=0 & \\
3x-4=0 & 
\end{cases}\\
\Rightarrow \begin{cases}
x=\frac{4}{3} & \\
x=y & 
\end{cases}
\end{array}$

Vậy $\displaystyle x=y=\frac{4}{3}$ là nghiệm pt