Question

Giải phương trình \(\sqrt{x^2+15}=3x-2+\sqrt{x^2+8}\).

Answer 1

\(3x-2=\sqrt[]{x^2+15}-\sqrt[]{x^2+8}=\dfrac{7}{\sqrt[]{x^2+15}+\sqrt[]{x^2+8}}>0\)

\(\Rightarrow x>\dfrac{2}{3}\)

\(\sqrt[]{x^2+15}-4=3x-3+\sqrt[]{x^2+8}-3\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt[]{x^2+15}+4}=3\left(x-1\right)+\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt[]{x^2+8}+3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{x+1}{\sqrt[]{x^2+15}+4}=3+\dfrac{x+1}{\sqrt[]{x^2+8}+3}\left(1\right)\end{matrix}\right.\)

Do \(x>\dfrac{2}{3}\Rightarrow x+1>0\Rightarrow\dfrac{x+1}{\sqrt[]{x^2+15}+4}< \dfrac{x+1}{\sqrt[]{x^2+8}+3}\)

\(\Rightarrow\) (1) vô nghiệm hay pt có nghiệm duy nhất \(x=1\)