Question

Giải bất phương trình: \(\frac{1-\sqrt{1-8x^2}}{2x}< 1\)

Answer 1

ĐKXĐ: \(\frac{-\sqrt{2}}{4}\le x\le\frac{\sqrt{2}}{4};x\ne0\)

\(\Leftrightarrow\frac{1-\sqrt{1-8x^2}-2x}{2x}< 0\)

TH1: \(\frac{-\sqrt{2}}{4}\le x< 0\)

\(\Rightarrow1-\sqrt{1-8x^2}-2x>0\)

\(\Rightarrow\sqrt{1-8x^2}>2x+1\)

Do \(x\ge-\frac{\sqrt{2}}{4}\Rightarrow2x+1>0\)

\(\Rightarrow1-8x^2>4x^2+4x+1\)

\(\Rightarrow12x^2+4x< 0\)

\(\Rightarrow\frac{-1}{3}< x< 0\)

TH2: \(0< x\le\frac{\sqrt{2}}{4}\)

\(\Rightarrow1-\sqrt{1-8x^2}-2x< 0\)

\(\Rightarrow2x+1>\sqrt{1-8x^2}\)

\(\Rightarrow12x^2+4x>0\)

\(\Rightarrow0< x\le\frac{\sqrt{2}}{4}\)

Vậy nghiệm của BPT là \(\left\{{}\begin{matrix}-\frac{1}{3}< x< 0\\0< x\le\frac{\sqrt{2}}{4}\end{matrix}\right.\)