5. \(10x^2+24xy-28x+16y^2-24y+41\)
\(=x^2-10x+25+9x^2+16y^2+9+24xy-18x-24y+7\)
\(=\left(x-5\right)^2+\left(3x+4y-3\right)^2+7\ge7\)
Dấu \(=\)khi \(\hept{\begin{cases}x-5=0\\3x+4y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=-3\end{cases}}\)