`~rai~`
\(a)\sqrt{4-5x}=12\quad(x\le\dfrac{4}{5})\\\Leftrightarrow (\sqrt{4-5x})^2=12^2\\\Leftrightarrow 4-5x=144\\\Leftrightarrow 5x=-140\\\Leftrightarrow x=-28.\text{(thỏa mãn ĐKXĐ)}\\\text{Vậy x=-28.}\\b)\sqrt{(2x-3)^2}=5\quad(x\ge\dfrac{3}{2}\\\Leftrightarrow |2x-3|=5\\\Leftrightarrow \left[\begin{array}{I}2x-3=5\\2x-3=-5\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=8\\2x=-2\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=4\text{(thỏa mãn)}\\x=-1\text{(không thỏa mãn)}\end{array}\right.\\c)\sqrt{x+3}+\sqrt{4x+12}=5\quad(x\ge -3)\\\Leftrightarrow \sqrt{x+3}+\sqrt{4.(x+3)}=5\\\Leftrightarrow \sqrt{x+3}+\sqrt{4}.\sqrt{x+3}=5\\\Leftrightarrow \sqrt{x+3}+2\sqrt{x+3}=5\\\Leftrightarrow 3\sqrt{x+3}=5\\\Leftrightarrow \sqrt{x+3}=\dfrac{5}{3}\\\Leftrightarrow (\sqrt{x+3})^2=\left(\dfrac{5}{3}\right)^2\\\Leftrightarrow x+3=\dfrac{25}{9}\\\Leftrightarrow x=-\dfrac{2}{9}.\text{(thỏa mãn ĐKXĐ)}\\\text{Vậy x=}-\dfrac{2}{9}.\)
