\(\sqrt{1+6x+9x^2}\) = 7
\(\Leftrightarrow\)\(\sqrt{\left(1+3x\right)^2}=7\)
\(\Leftrightarrow\left|1+3x\right|=7\)
\(\Leftrightarrow1+3x=7\)
\(\Leftrightarrow3x=7-1=6\)
\(\Leftrightarrow x=\frac{6}{3}=2\)
Vậy x=2
ĐKXĐ: \(x\in R\)
PT\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=7\Leftrightarrow\left|3x+1\right|=7\)
Nếu \(x\ge\frac{-1}{3}\)thì \(3x+1=7\Leftrightarrow x=2\)(thỏa mãn)
Nếu \(x< \frac{-1}{3}\)thì \(-3x-1=7\Leftrightarrow x=\frac{-8}{3}\)(thỏa mãn)
