ĐKXĐ : Luôn đúng .
Ta có : \(\sqrt{4x^2+4x+1}=6\)
=> \(\sqrt{\left(2x+1\right)^2}=6\)
=> \(\left|2x+1\right|=6\)
=> \(\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
Vậy ...
ĐKXĐ: \(x\in R\)
Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\left(nhận\right)\\x=\frac{-7}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
\(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(4x+1\right)^2}=6\)
\(\Leftrightarrow\left|4x+1\right|=6\)
\(\Leftrightarrow4x+1=6\)
\(\Leftrightarrow4x=6-1\)
\(\Leftrightarrow4x=5\)
\(\Leftrightarrow x=\frac{5}{4}\)
Vậy S = \(\left\{\frac{5}{4}\right\}\)
