Question

Tìm x

 a)\(\sqrt{\left(2x-1\right)^2}=x+1\)

b)\(\sqrt{x+3}=5\)

c)\(\sqrt{x+2}=\sqrt{7}\)

Answer 1

b: Ta có: \(\sqrt{x+3}=5\)

\(\Leftrightarrow x+3=25\)

hay x=22

c: Ta có: \(\sqrt{x+2}=\sqrt{7}\)

\(\Leftrightarrow x+2=7\)

hay x=5

Answer 2

a:Ta có: \(\sqrt{\left(2x-1\right)^2}=x+1\)

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

\(\Leftrightarrow\left(x-2\right)\cdot3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)