Question

giải bất phương trình giúp mình với ạ

\(\frac{4}{\sqrt{x}}+\sqrt{2x+1}\ge\sqrt{2x+17}\)

Answer 1

ĐKXĐ: \(x>0\)

\(\Leftrightarrow\frac{16}{x}+2x+1+8\sqrt{\frac{2x+1}{x}}\ge2x+17\)

\(\Leftrightarrow\frac{2}{x}+\sqrt{2+\frac{1}{x}}-2\ge0\)

Đặt \(\sqrt{2+\frac{1}{x}}=t>0\Rightarrow\frac{1}{x}=t^2-2\)

\(2\left(t^2-2\right)+t-2\ge0\)

\(\Leftrightarrow2t^2+t-6\ge0\) \(\Rightarrow\left[{}\begin{matrix}t\ge\frac{3}{2}\\t\le-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2+\frac{1}{x}}\ge\frac{3}{2}\Rightarrow\frac{1}{x}\ge\frac{1}{4}\Rightarrow x\le4\)

\(\Rightarrow0< x\le4\)