a. $(x+\frac{5}{3}).(x-$ $\frac{5}{4})=0$
⇔ \(\left[ \begin{array}{l}x+\frac{5}{3}=0\\x-\frac{5}{4}=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-5}{3}\\x=\frac{5}{4}\end{array} \right.\)
Vậy $x=\frac{-5}{3}$ hoặc $x=\frac{5}{4}$
b. $(x-\frac{1}{2})^2=4$
⇔ \(\left[ \begin{array}{l}x-\frac{1}{2}=2\\x-\frac{1}{2}=-2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2+\frac{1}{2}\\x=-2+\frac{1}{2}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{-3}{2}\end{array} \right.\)
Vậy $x=\frac{5}{2}$ hoặc $x=\frac{-3}{2}$
c. $(x+1)^3=-125$
⇔ $(x+1)^3=(-5)^3$
⇔ $x+1=-5$
⇔ $x=-5-1$
⇔ $x=-6$
Vậy $x=-6$
