Question

Answer 1

a: Ta có: \(3\sqrt{x}+2=\sqrt{x}+10\)

\(\Leftrightarrow2\sqrt{x}=8\)

\(\Leftrightarrow\sqrt{x}=4\)

hay x=16

b: Ta có: \(\sqrt{x^2+x+1}=1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)