Question

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

giúp tui vs 

Answer 1

\(\sqrt{x^2+2}=\sqrt{1-2x}\left(đkxđ:x\le\dfrac{1}{2}\right)\)

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

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

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

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\left(tmđk\right)\)

Answer 2

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

=>\(\left\{{}\begin{matrix}1-2x>=0\\x^2+2=1-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\x^2+2x+1=0\end{matrix}\right.\)

=>x=-1