\(ĐKXĐ:x\ge-4 \\pt<=>9x^2-6x+1=x^2+8x+16 \\<=>8x^2-14x-15=0 \\<=>x=\dfrac{5}{2}\ (thoả\ ĐKXĐ)\ hoặc\ x=\dfrac{-3}{4} (thoả\ ĐKXĐ)\)
\(đkx\ge-4\\ \sqrt{\left(3x-1\right)^2}=x+4\\ \left|3x-1\right|=x+4\\ 3x-1=x+4\\ 3x-x=4+1\\ 2x=5\\ x=\dfrac{5}{2}\left(thoaman\right)\)
`3x - 1 = x + 4`
`2x = 5`
`->x = 2,5`
\(\sqrt{9x^2-6x+1}=x+4\)
ĐK: \(x\ge-4\)
\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}=x+4\)
\(\Leftrightarrow\left|3x-1\right|=x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+4\\3x+1=-x-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{4}\end{matrix}\right.\)