a) \(\Leftrightarrow\sqrt{\left(4x-7\right)^2}=5\)
\(\Leftrightarrow\left|4x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-7=5\\4x-7=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(đk:x\ge-\dfrac{5}{6}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=6x+5\)
\(\Leftrightarrow\left|2x-1\right|=6x+5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=6x+5\left(x\ge\dfrac{1}{2}\right)\\2x-1=-6x-5\left(-\dfrac{5}{6}\le x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(ktm\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)
c) \(đk:x\ge3\)
\(\Leftrightarrow6\sqrt{x-3}+2\sqrt{x-3}-3\sqrt{x-3}=15\)
\(\Leftrightarrow5\sqrt{x-3}=15\Leftrightarrow\sqrt{x-3}=3\)
\(\Leftrightarrow x-3=9\Leftrightarrow x=12\left(tm\right)\)
d) \(đk:x\ge-3\)
\(\Leftrightarrow3\sqrt{x+3}+\sqrt{x+3}-\sqrt{x+3}=12\)
\(\Leftrightarrow3\sqrt{x+3}=12\Leftrightarrow\sqrt{x+3}=4\)
\(\Leftrightarrow x+3=16\Leftrightarrow x=13\left(tm\right)\)
e) \(đk:x\ge3\)
\(\Leftrightarrow4\sqrt{x-3}-3\sqrt{x-3}+2\sqrt{x-3}=5\)
\(\Leftrightarrow3\sqrt{x-3}=5\Leftrightarrow\sqrt{x-3}=\dfrac{5}{3}\)
\(\Leftrightarrow x-3=\dfrac{25}{9}\Leftrightarrow x=\dfrac{52}{9}\left(tm\right)\)
a: Ta có: \(\sqrt{16x^2-56x+49}=5\)
\(\Leftrightarrow\left|4x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-7=5\\4x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=12\\4x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
e: Ta có: \(\sqrt{16x-48}-6\sqrt{\dfrac{x-3}{4}}+\sqrt{4x-12}=5\)
\(\Leftrightarrow4\sqrt{x-3}-3\sqrt{x-3}+2\sqrt{x-3}=5\)
\(\Leftrightarrow\sqrt{x-3}=\dfrac{5}{3}\)
\(\Leftrightarrow x-3=\dfrac{25}{9}\)
hay \(x=\dfrac{52}{9}\)
