Question

Tìm GTNN

a) C=2x²+x-15

b) D=3x²+10x+20

Answer 1

Lời giải:

a)

\(C=2x^2+x-15=2(x^2+\frac{x}{2}+\frac{1}{4^2})-\frac{121}{8}\)

\(=2(x+\frac{1}{4})^2-\frac{121}{8}\)

Vì \((x+\frac{1}{4})^2\geq 0, \forall x\Rightarrow C\geq 2.0-\frac{121}{8}=-\frac{121}{8}\)

Vậy \(C_{\min}=\frac{-121}{8}\Leftrightarrow x=-\frac{1}{4}\)

b) Ta có:

\(D=3x^2+10x+20=3(x^2+\frac{10}{3}x+\frac{5^2}{3^2})+\frac{35}{3}\)

\(=3(x+\frac{5}{3})^2+\frac{35}{3}\)

Vì \((x+\frac{5}{3})^2\geq 0, \forall x\in\mathbb{R}\) \(\Rightarrow D\geq 3. 0+\frac{35}{3}=\frac{35}{3}\)

Vậy \(D_{\min}=\frac{35}{3}\Leftrightarrow x=-\frac{5}{3}\)