Question

[Sách Giáo Khoa]

Tìm \(x\), biết :

a) \(\sqrt{\left(2x-1\right)^2}=3\)

b) \(\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\dfrac{1}{3}\sqrt{15x}\)

Answer 1

[Lưu Hạ Vy]

Answer 2

[Đoàn Như Quỳnhh]

a) \(\sqrt{\left(2x-1\right)^2}=3\)

\(\Leftrightarrow\left|2x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\dfrac{1}{3}\sqrt{15x}\)

\(\Leftrightarrow\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\)

\(\Leftrightarrow\sqrt{15x}=6\)

\(\Leftrightarrow15x=6^2\Leftrightarrow15x=36\)

\(\Rightarrow x=\dfrac{5}{12}\)