a) \(x^2-4xy+5y^2+10x-22y+100\)
\(=\left(x^2+4y^2+25+10x-4xy-20y\right)+\left(y^2-2y+1\right)+74\)
\(=\left[\left(x-2y\right)^2+10\left(x-2y\right)+25\right]+\left(y-1\right)^2+74\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+74\)
- Vì \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+74\ge74>0\)
hay \(x^2-4xy+5y^2+10x-22y+100>0\)
- Mà tại \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\) thì biểu thức đạt GTNN là 74.
\(\Rightarrowđpcm\).
b) \(5x^2+10y^2-6xy-4x-2y+3\)
\(=5\left(x^2+\dfrac{9}{25}y^2+\dfrac{4}{25}-\dfrac{6}{5}xy-\dfrac{4}{5}x+\dfrac{12}{25}y\right)+\dfrac{41}{5}\left(y^2-\dfrac{22}{41}y+\dfrac{11^2}{41^2}\right)+\dfrac{66}{41}\)
\(=5\left[\left(x-\dfrac{3}{5}y\right)^2-\dfrac{4}{5}\left(x-\dfrac{3}{5}y\right)+\dfrac{4}{25}\right]+\dfrac{41}{5}\left(y-\dfrac{11}{41}\right)^2+\dfrac{66}{41}\)
\(=5\left(x-\dfrac{3}{5}y-\dfrac{2}{5}\right)^2+\dfrac{41}{5}\left(y-\dfrac{11}{41}\right)^2+\dfrac{66}{41}\)
- Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{3}{5}y-\dfrac{2}{5}\right)^2\ge0\\\left(y-\dfrac{11}{41}\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow5\left(x-\dfrac{3}{5}y-\dfrac{2}{5}\right)^2+\dfrac{41}{5}\left(y-\dfrac{11}{41}\right)^2+\dfrac{66}{41}\ge\dfrac{66}{41}\)
hay \(5x^2+10y^2-6xy-4x-2y+3\ge\dfrac{66}{41}>0\)
- Vì tại \(\left\{{}\begin{matrix}x=\dfrac{23}{41}\\y=\dfrac{11}{41}\end{matrix}\right.\), biểu thức đạt GTNN là \(\dfrac{66}{41}\).
\(\Rightarrowđpcm\)