\(d,\Leftrightarrow\sqrt{\left(x-\dfrac{1}{4}\right)^2}=\dfrac{1}{4}+1=\dfrac{5}{4}\\\Leftrightarrow\left|x-\dfrac{1}{4}\right|=\dfrac{5}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{5}{4},\forall x-\dfrac{1}{4}\ge0\\\dfrac{1}{4}-x=\dfrac{5}{4},\forall x-\dfrac{1}{4}< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2},\forall x\ge\dfrac{1}{4}\left(tm\right)\\x=-1,\forall x< \dfrac{1}{4}\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right. \)
\(e,ĐK:x\ge3\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)
\(f,ĐK:x\ge\dfrac{1}{2}\\ \Leftrightarrow\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tm\right)\)
\(g,2\sqrt{7-3x}+1=9\left(x\le\dfrac{7}{3}\right)\\ \Leftrightarrow\sqrt{7-3x}=4\Leftrightarrow7-3x=16\Leftrightarrow x=-3\left(tm\right)\)
