Question

Cho \(x,y\in R\)

Và \(\sqrt{x^2+5}+\sqrt{x-1}+x^2=\sqrt{y^2+5}+\sqrt{y-1}+y^2\)

C/m x=y

Answer 1

\(\sqrt{x^2+5}+\sqrt{x-1}+x^2=\sqrt{y^2+5}+\sqrt{y-1}+y^2\)

\(\Leftrightarrow\left(\sqrt{x^2+5}-\sqrt{y^2+5}\right)+\left(\sqrt{x-1}-\sqrt{y-1}\right)+\left(x^2-y^2\right)=0\)

\(\Leftrightarrow\frac{x^2-y^2}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}+\left(x^2-y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{x+y}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\frac{1}{\sqrt{x-1}+\sqrt{y-1}}+x+y\right)=0\)

\(\Leftrightarrow x=y\)