Bài 4:
a: Ta có: \(\sqrt{\left(2x-1\right)^2}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b: Ta có: \(\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\dfrac{1}{3}\sqrt{15x}\)
\(\Leftrightarrow\sqrt{15x}\cdot\dfrac{1}{3}=2\)
\(\Leftrightarrow15x=36\)
hay \(x=\dfrac{12}{5}\)
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