1. \(\sqrt{1-12x+36x^2}=5\)
ĐKXD: \(1-12x+36x^2\ge0\Leftrightarrow\forall x\in R\)
\(\Leftrightarrow\sqrt{\left(6x\right)^2-2.6x.1+1^2}=5\)
\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2. \(\sqrt{1-4x+4x^2}=5\)
ĐKXD: \(1-4x+4x^2\ge0\Leftrightarrow\forall x\in R\)
\(\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.1+1^2}=5\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)
\(\Leftrightarrow\left|2x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
4. \(\sqrt{x^2+6x+9}=x-1\)
ĐKXD: \(x^2+6x+9\ge0\Leftrightarrow\forall x\in R\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=x-1\)
\(\Leftrightarrow\left|x+3\right|=x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x-1\\x+3=-x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=-4\\2x=-2\end{matrix}\right.\Leftrightarrow x=-1\)
5, \(\sqrt{x^2+x+\dfrac{1}{4}}=x\)
ĐKXD: \(x^2+x+\dfrac{1}{4}\ge0\Leftrightarrow\forall x\in R\)
\(\Leftrightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2}=x\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=x\\x+\dfrac{1}{2}=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=-\dfrac{1}{2}\\2x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{4}\)
Câu 3 thiếu đề nhé.
