Question

Answer 1

\(\sqrt{x^2-2x-8}=\sqrt{3}\left(x-1\right)\left(đk:x\ge4\right)\)

\(\Leftrightarrow x^2-2x-8=3\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x-8=3x^2-6x+3\)

\(\Leftrightarrow2x^2-4x+11=0\) (vô lý do \(2x^2-4x+11=2\left(x-1\right)^2+9\ge9>0\))

Vậy \(S=\varnothing\)

Answer 2

\(ĐK:x\le-2;x\ge4\\ \Leftrightarrow x^2-2x-8=3\left(x-1\right)^2\\ \Leftrightarrow x^2-2x-8=3x^2-6x+3\\ \Leftrightarrow2x^2-4x+11=0\\ \Leftrightarrow2\left(x-1\right)^2+9=0\Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+9\ge9>0\right]\)