Question

Giải bpt sau : \(x^2 - 4x - 6 - \sqrt{2x^2 - 8x + 12} \geq 0\)

Answer 1

Đặt \(\sqrt{2x^2-8x+12}=t>0\)

\(\Rightarrow x^2-4x=\frac{t^2-12}{2}\)

BPT trở thành:

\(\frac{t^2-12}{2}-6-t\ge0\)

\(\Leftrightarrow t^2-2t-24\ge0\Rightarrow\left[{}\begin{matrix}t\ge6\\t\le-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2-8x+12}\ge6\)

\(\Leftrightarrow2x^2-8x+12\ge36\)

\(\Leftrightarrow x^2-4x-12\ge0\Rightarrow\left[{}\begin{matrix}x\ge6\\x\le-2\end{matrix}\right.\)