\(x^2+x+1=\left(x^2+\frac{1}{2}\cdot2\cdot x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra khi \(x=-\frac{1}{2}\)
\(4x^2+4x-5=\left(4x^2+4x+1\right)-6=\left(2x+1\right)^2-6\ge-6\)
Dấu "=" xảy ra khi \(x=-\frac{1}{2}\)
\(\left(x-3\right)\left(x+5\right)+4\)
\(=x^2+2x+19\)
\(=\left(x^2+2x+1\right)+18\)
\(=\left(x+1\right)^2+18\)
\(\ge18\)
Dấu "=" xảy ra khi \(x=-1\)
\(x^2-4x+y^2-8y+6\)
\(=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\)
\(\ge-14\)
Dấu "=" xảy ra khi \(x=2;y=4\)
