a)
\(\sqrt{9x^2}=12\\ < =>\left(\sqrt{9x^2}\right)^2=12^2\\ < =>9x^2=144\\ < =>x^2=16\\ < =>\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
b)
\(\sqrt{25x^2}=\left|-50\right|\\ < =>\sqrt{25x^2}=50\left(vì-50< 0\right)\)
\(< =>\left(\sqrt{25x^2}\right)^2=50^2\\ =>25x^2=2500\\ < =>x^2=100\\ < =>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
\(a)\sqrt{9x^2}=12\\ \Leftrightarrow\sqrt{\left(3x\right)^2}=12\\ \Leftrightarrow\left|3x\right|=12\\ \Leftrightarrow\left|x\right|=\dfrac{12}{3}\\ \Leftrightarrow x=4\)
Vậy x=4
\(b)\sqrt{25x^2}=\left|-50\right|\\ \Leftrightarrow\sqrt{\left(5x\right)^2}=\left|-50\right|\\ \Leftrightarrow\left|5x\right|=50\\ \Leftrightarrow\left|x\right|=\dfrac{50}{5}\\ \Leftrightarrow x=10\)
Vậy x=10
