Câu 74 Trang 40: Tìm x

a.  $\sqrt{(2x-1)^{2}}=3$

<=> $\left | 2x-1 \right |=3$

<=> $\left\{\begin{matrix}3\geq 0 &  & \\ 2x-1=3 &  & \\ 2x-1=-3 &  & \end{matrix}\right.$

<=> $\left\{\begin{matrix}2x=4 & \\ 2x=-2 & \end{matrix}\right.<=> \left\{\begin{matrix}x=2 & \\ x=-1 & \end{matrix}\right.$

Vậy $\left\{\begin{matrix}x=2 & \\ x=-1 & \end{matrix}\right.$

b.  $\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}$

<=> $\frac{5}{3}\sqrt{15x}-\sqrt{15x}-\frac{1}{3}\sqrt{15x}=2$

<=> $\sqrt{15x}(\frac{5}{3}-1-\frac{1}{3})=2$

<=> $\sqrt{15x}\frac{1}{3}=2<=> \sqrt{15x}=6$

<=> $15x=6^{2}<=> x=\frac{36}{15}=\frac{12}{5}$

Vậy $x=\frac{12}{5}$