a) \(\sqrt{\left(x+1\right)^2}=4\Rightarrow\left|x-1\right|=4\Rightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
còn câu b nãy mình làm rồi bạn xem lại
a) Ta có: \(\sqrt{\left(x+1\right)^2}=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x-20}-3\cdot\sqrt{\dfrac{x-5}{9}}=2\)
\(\Leftrightarrow2\sqrt{x-5}-\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
hay x=9
