Không có ai trả lời thì cho mình vậy :))
\(\sqrt{x+4}\sqrt{x-4}=2x-12+2\sqrt{x^2-16}\)
\(\Rightarrow\sqrt{\left(x+4\right)\left(x-4\right)}=2x-12+2\sqrt{x^2-16}\)
\(\Leftrightarrow\sqrt{x^2-16}=2x-12+2\sqrt{x^2-16}\)
\(\Leftrightarrow\sqrt{x^2-16}-2\sqrt{x^2-16}=2x-12\)
\(\Leftrightarrow-\sqrt{x^2-16}=2x-12\)
\(\Leftrightarrow\sqrt{x^2-16}=-2x+12\)
\(\Leftrightarrow x^2-16=\left(-2x+12\right)^2\)
\(\Leftrightarrow x^2-16=4x^2-48x+144\)
\(\Leftrightarrow x^2-16-4x^2+48x-144=0\)
\(\Leftrightarrow-3x^2-160+48x=0\)
\(\Leftrightarrow-3x^2+48x-160=0\)
\(\Leftrightarrow3x^2-48x+160=0\)
\(\Leftrightarrow x=\dfrac{-\left(-48\right)\pm\sqrt{\left(-48\right)^2-4\cdot3\cdot160}}{2\cdot3}\)
\(\Leftrightarrow x=\dfrac{48\pm\sqrt{2304-1920}}{6}\)
\(\Leftrightarrow x=\dfrac{48\pm\sqrt{384}}{6}\)
\(\Leftrightarrow x=\dfrac{48+8\sqrt{6}}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{48+8\sqrt{6}}{6}\\x=\dfrac{48-8\sqrt{6}}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{24+4\sqrt{6}}{3}\\x=\dfrac{24-4\sqrt{6}}{3}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{24+4\sqrt{6}}{3};x_2=\dfrac{24-4\sqrt{6}}{3}\)
