Question

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

Giải phương trình

Answer 1

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1}+\sqrt{\left(x-1\right)\left(x+1\right)}=x\sqrt{x}\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}+1\right)=x\sqrt{x}\)

\(\Leftrightarrow\dfrac{\sqrt{x-1}.x}{\sqrt{x+1}-1}=x\sqrt{x}\)

\(\Leftrightarrow\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}=\sqrt{x}\)

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

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

\(\Leftrightarrow2x-1+2\sqrt{x^2-x}=x^2+x\)

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

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

\(\Leftrightarrow x^2-x-1=0\)