Question

giải phương trình

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

Answer 1

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)( t/m )