\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)
\(\Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)
\(\left(ĐK:x>\dfrac{5}{3}\right)\)
\(\Leftrightarrow\left(\sqrt{x^2+12}-4\right)+5=3x+\left(\sqrt{x^2+5}-3\right)-1\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2+12}+4}-\dfrac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2+5}+3}+3\left(2-x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{x+2}{\sqrt{x^2+12}+4}-\dfrac{x+2}{\sqrt{x^2+5}+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\dfrac{x+2}{\sqrt{x^2+12}+4}-\dfrac{x+2}{\sqrt{x^2+5}+3}-3=0\end{matrix}\right.\)
Ta có: \(\sqrt{x^2+12}+4>\sqrt{x^2+5}+3\)
\(\Rightarrow\dfrac{x+2}{\sqrt{x^2+12}+4}< \dfrac{x+2}{\sqrt{x^2+5}+3}\)
\(\Rightarrow\dfrac{x+2}{\sqrt{x^2+12}+4}-\dfrac{x+2}{\sqrt{x^2+5}+3}< 0\)
Suy ra: \(\dfrac{x+2}{\sqrt{x^2+12}+4}-\dfrac{x+2}{\sqrt{x^2+5}+3}-3=0\left(loai\right)\)
\(\Rightarrow x-2=0\)
\(\Leftrightarrow x=2\)
