Question

giúp mình câu 3,4,8 

Answer 1

1: Ta có: \(\sqrt{x^2+6x+9}=3x-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=3x-1\left(x\ge-3\right)\\x+3=1-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3x=-1-3\\x+3x=1-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=-4\\4x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)

2: Ta có: \(\sqrt{1-4x+4x^2}=5\)

\(\Leftrightarrow\left|2x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

3: Ta có: \(\sqrt{9x^2}=2x+1\)

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

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

Answer 2

4: Ta có: \(\sqrt{x^4}=7\)

\(\Leftrightarrow x^2=7\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

8: Ta có: \(\sqrt{2x-1}=\sqrt{5}\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

hay x=3