Question

a ) Chứng tỏ rằng : \(x-2\sqrt{x}+17>0\) với mọi \(x\ge0\)

b ) Tìm giá trị nhỏ nhất của \(x-5\sqrt{x}-2018\)

Answer 1

a/ \(x-2\sqrt{x}+1+16=\left(\sqrt{x}-1\right)^2+16>0\forall x\ge0\)

b/ \(x-5\sqrt{x}+\frac{25}{4}-\frac{8097}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{8097}{4}\ge-\frac{8097}{4}\)

"="\(\Leftrightarrow x=\frac{25}{4}\)

Answer 2

a) \(x-2\sqrt{x}+17\)

\(=x-2\sqrt{x}+1+16\)

\(=\left(\sqrt{x}-1\right)^2+16>0\forall x\ge0\)

b) \(x-5\sqrt{x}-2018\)

\(=x-2\sqrt{x}\cdot\frac{5}{2}+\frac{25}{4}-\frac{8097}{4}\)

\(=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{8097}{4}\ge-\frac{8097}{4}\forall x\ge0\)

Dấu "=" \(\Leftrightarrow\sqrt{x}=\frac{5}{2}\Leftrightarrow x=\frac{25}{4}\)