a) \(x^2-8x+20\)
\(=x^2-2.x.4+16+4\)
\(=\left(x-4\right)^2+4\)
Có: \(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2+4>0\)
Hay:.............
b) \(x^2+11\)
Có: \(x^2\ge0\Rightarrow x^2+11>0\)
Hay:.............
c) \(4x^2-12x+11\)
\(=4\left(x^2-3x+\frac{11}{4}\right)\)
\(=4\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{1}{2}\right)\)
\(=4\left(x-\frac{3}{2}\right)^2+2>0\)
d) \(x^2+5y^2+2x+6y+34\)
\(=x^2+2.x.1+1+y^2+4y^2+2.y.3+9+24\)
\(=\left(x^2+2.x.1+1\right)+\left(y^2+2.y.3+9\right)+4y^2+24\)
\(=\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24\)
Ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\\\left(2y\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24>0\)
f) \(x^2-2x+y^2+4y+6\)
\(=x^2-2.x.1+1+y^2+2.y.2+4+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1>0\)
