a) \(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\left(x\ge\dfrac{1}{6}\right)\\6x-1=-5\left(x< \dfrac{1}{6}\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
b) \(\sqrt{x^2-25}-\sqrt{x-5}=0\left(đk:x\ge5\right)\)
\(\Leftrightarrow\sqrt{x^2-25}=\sqrt{x-5}\)
\(\Leftrightarrow x^2-25=x-5\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-4\left(l\right)\end{matrix}\right.\)
d) \(\sqrt{x^2-4}+\sqrt{x-2}=0\left(đk:x\ge2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)( do \(\sqrt{x^2-4},\sqrt{x-2}\ge0\))
\(\Leftrightarrow x=2\left(tm\right)\)
e) \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đk:x\ge5\right)\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
f: Ta có: \(3x-7\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{16}{9}\end{matrix}\right.\)
