Question

Answer 1

a: Ta có: \(\sqrt{3x+1}=4\)

\(\Leftrightarrow3x+1=16\)

\(\Leftrightarrow3x=15\)

hay x=5

b: Ta có: \(\sqrt{x^2-10x+25}=12\)

\(\Leftrightarrow\left|x-5\right|=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=12\\x-5=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-7\end{matrix}\right.\)

Answer 2

c: Ta có: \(3\sqrt{4x+8}-\sqrt{25x+50}+\sqrt{x+2}=2\)

\(\Leftrightarrow6\sqrt{x+2}-5\sqrt{x+2}+\sqrt{x+2}=2\)

\(\Leftrightarrow2\sqrt{x+2}=2\)

\(\Leftrightarrow x+2=1\)

hay x=-1