Question

\(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)

Answer 1

ĐKXĐ: ...

\(\sqrt{2x^2+2x+3}-\sqrt{2x^2-1}+\sqrt{x^2-x+2}-\sqrt{x^2-3x-2}=0\)

\(\Leftrightarrow\frac{2x^2+2x+3-\left(2x^2-1\right)}{\sqrt{2x^2+2x+3}+\sqrt{2x^2-1}}+\frac{x^2-x+2-\left(x^2-3x-2\right)}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}=0\)

\(\Leftrightarrow\frac{2\left(x+2\right)}{\sqrt{2x^2+2x+3}+\sqrt{2x^2-1}}+\frac{2\left(x+2\right)}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}=0\)

\(\Leftrightarrow\left(x+2\right)\left(\frac{2}{\sqrt{2x^2+2x+3}+\sqrt{2x^2-1}}+\frac{2}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)