a) \(\sqrt{\left(x-3\right)^2}=3\Leftrightarrow\left|x-3\right|=3\) \(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
b) \(\sqrt{4x^2-20x+25}+2x=5\Leftrightarrow\left|2x-5\right|+2x-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x-5+2x-5=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\5-2x+2x-5=0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x-10=0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\0x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x=\dfrac{10}{4}\left(N\right)\end{matrix}\right.\\x\le\dfrac{5}{2}\end{matrix}\right.\) ** 10/4 = 5/2 rồi**
Kl: x \< 5/2
c) \(\sqrt{1-12x+36x^2}=5\Leftrightarrow\left|1-6x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-6x=5\\1-6x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(N\right)\\x=1\left(N\right)\end{matrix}\right.\)
Kl: x=-2/3, x=1
d) Đk: x >/ 1
\(\sqrt{x+2\sqrt{x-1}}=2\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}+1=2\left(1\right)\\\sqrt{x-1}+2=-2\left(VN\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\)(N)
Kl: x=2
e) Đk: x >/ 1
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge1\\\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}-1=\sqrt{x-1}-1\) (luôn đúng)
kl: x >/ 1
f) \(\sqrt{x^2-\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\left|\dfrac{1}{4}-x\right|=\dfrac{1}{4}-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\dfrac{1}{4}-x=\dfrac{1}{4}-x\end{matrix}\right.\)
(luôn đúng)
Kl: x \< 1/4
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