Question

Giải phương trình:

\(\sqrt{2x-1}=\sqrt{5}\) và \(\sqrt{9x^2}=2x+1\)

Answer 1

a: Ta có: \(\sqrt{2x-1}=\sqrt{5}\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

hay x=3

b: Ta có: \(\sqrt{9x^2}=2x+1\)

\(\Leftrightarrow\left|3x\right|=2x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\left(x\ge0\right)\\-3x=2x+1\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{-1}{5}\left(nhận\right)\end{matrix}\right.\)