Question

giải bpt

a) \(x^2-3x-\sqrt{x^2-3x+5}>1\)

b) \(\sqrt[4]{x-\sqrt{x^2-1}}+4\sqrt{x+\sqrt{x^2-1}}-3< 0\)

Answer 1

a/ Đặt \(\sqrt{x^2-3x+5}=t>0\)

\(\Leftrightarrow t^2-5-t>1\Leftrightarrow t^2-t-6>0\)

\(\Rightarrow\left[{}\begin{matrix}t>3\\t< -2\left(l\right)\end{matrix}\right.\) \(\Rightarrow\sqrt{x^2-3x+5}>3\)

\(\Leftrightarrow x^2-3x+5>9\Leftrightarrow x^2-3x-4>0\Rightarrow\left[{}\begin{matrix}x>4\\x< -1\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ge1\)

Đặt \(\sqrt[4]{x-\sqrt{x^2-1}}=t>0\Rightarrow\sqrt[4]{x+\sqrt{x^2-1}}=\frac{1}{t}\)

\(\Leftrightarrow t+\frac{4}{t^2}-3< 0\)

\(\Leftrightarrow t^3-3t^2+4< 0\)

\(\Leftrightarrow\left(t+1\right)\left(t-2\right)^2< 0\)

Do \(t>0\Rightarrow t+1>0\Rightarrow VT\ge0\Rightarrow\) BPT vô nghiệm