Bài 1:
b) Ta có: \(B=x^2+8x+20\)
\(=x^2+8x+16+4\)
\(=\left(x+4\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=-4
c) Ta có: \(C=x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
Bài 2:
b) Ta có: \(B=-x^2-8x-20\)
\(=-\left(x^2+8x+20\right)\)
\(=-\left(x+4\right)^2-4\le-4\forall x\)
Dấu '=' xảy ra khi x=-4
c) Ta có: \(C=-x^2+x-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
