Question

help em cau 3 vs

 

Answer 1

Câu 2: 

a) Ta có: \(\sqrt{9x^2-12x+4}-7=0\)

\(\Leftrightarrow\left|3x-2\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=7\\3x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) Ta có: \(3\sqrt{4x-12}=15+\dfrac{1}{3}\sqrt{9x-27}\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}=15\)

\(\Leftrightarrow5\sqrt{x-3}=15\)

\(\Leftrightarrow\sqrt{x-3}=3\)

\(\Leftrightarrow x-3=9\)

hay x=12