\(4x^2+4x-5\)
\(=4x^2+4x+1-6\)
\(=\left(2x+1\right)^2-6\ge-6\)
\(\Rightarrow Min=-6\Leftrightarrow x=-\dfrac{1}{2}\)
\(\left(x-3\right)\left(x+5\right)+4\)
\(=x^2+5x-3x-15+4\)
\(=x^2+2x-11\)
\(=x^2+2x-1-10\)
\(=\left(x-1\right)^2-10\ge-10\)
\(\Rightarrow Min=-10\Leftrightarrow x=1\)
\(x^2-4x+y^2-8y+6\)
\(=\left(x^2-4x+4\right)+\left(y^2-8x+16\right)-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(\Rightarrow Min=-14\Leftrightarrow x=2;y=4\)
a: Ta có: \(4x^2+4x-5\)
\(=4x^2+4x+1-6\)
\(=\left(2x+1\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x-3\right)\left(x+5\right)+4\)
\(=x^2+2x-15+4\)
\(=x^2+2x+1-12\)
\(=\left(x+1\right)^2-12\ge-12\forall x\)
Dấu '=' xảy ra khi x=-1
c: Ta có: \(x^2-4x+y^2-8y+6\)
\(=x^2-4x+4+y^2-8y+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x,y\)
Dấu '=' xảy ra khi x=2 và y=4
