1)\(\sqrt{-3x+4}=12\left(đk:x\le\dfrac{4}{3}\right)\Leftrightarrow-3x+4=144\Leftrightarrow-3x=140\Leftrightarrow x=-\dfrac{140}{3}\)( thỏa đk)
2) \(\sqrt{2x^2-9}=-x\left(đk:x\le-\dfrac{3\sqrt{2}}{2}\right)\)
\(\Rightarrow2x^2-9=x^2\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\Rightarrow x=-3\)( do đk)
3) \(\sqrt{\dfrac{12x+5}{3}}=2\left(đk:x\ge-\dfrac{5}{12}\right)\)
\(\Rightarrow\dfrac{12x+5}{3}=4\Rightarrow12x+5=12\Rightarrow x=\dfrac{7}{12}\)
1: Ta có: \(\sqrt{-3x+4}=12\)
\(\Leftrightarrow-3x+4=144\)
\(\Leftrightarrow-3x=140\)
hay \(x=-\dfrac{140}{3}\)
2: Ta có: \(\sqrt{2x^2-9}=-x\)
\(\Leftrightarrow2x^2-9=x^2\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=-3\left(x\le0\right)\)
3: Ta có: \(\sqrt{\dfrac{12x+5}{3}}=2\)
\(\Leftrightarrow12x+5=12\)
\(\Leftrightarrow12x=7\)
hay \(x=\dfrac{7}{12}\)
