\(\sqrt{x^2+6x+9}-1=2x\)
\(\sqrt{\left(x^2+2.x.3+3^2\right)}-1=2x\)
\(\sqrt{\left(x+3\right)^2}-1=2x\)
\(x+3-1=2x\)
\(x+2=2x\)
\(x=2\)
\(\Leftrightarrow\sqrt{x^2+6x+9}=2x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1\ge0\\x^2+6x+9=\left(2x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x^2+6x+9=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\3x^2-2x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=2\)
\(\sqrt{4x^2-4x+1}+1=3x\)
<=> \(\sqrt{\left(2x-1\right)^2}+1=3x\) ĐKXĐ: Mọi x
<=> \(|2x-1|+1=3x\)
<=> \(\left[{}\begin{matrix}2x-1+1=3x\\-\left(2x-1\right)+1=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3x\\-2x+1+1=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-3x=0\\-2x+2=3x\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x\left(2-3\right)=0\\-2x-3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-5x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(\sqrt{9x+9}-2\sqrt{\dfrac{x+1}{4}}=4\)
\(\Leftrightarrow2\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=4\)
hay x=3
