a/
$\frac{2}{3}$ + $\frac{7}{4}$ : $x$ =$\frac{5}{6}$ ⇔ $\frac{7}{4}$ : $x$ =$\frac{5}{6}$ -$\frac{2}{3}$ ⇔$\frac{7}{4}$ : $x$=$\frac{1}{6}$ ⇔$x$=$\frac{7}{4}$ : $\frac{1}{6}$ ⇔ $x$=$\frac{7}{4}$ .$6$⇔$x$=$\frac{14}{3}$
vậy x=14/3
b/
(x+$\frac{5}{3}$ )(x-$\frac{5}{4}$ )=0 ⇔\(\left[ \begin{array}{l}x+\frac{5}{3}=0\\x-\frac{5}{4}=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0-\frac{5}{3}\\x=0-\frac{5}{4}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-\frac{5}{3}\\x=-\frac{5}{4}\end{array} \right.\)
c/
$(x-1,2)^{2}$ =4⇔\(\left[ \begin{array}{l}x-1,2=2\\x-1,2=-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=3,2\\x=-0,8\end{array} \right.\)
d/
$(x+1)^{3}$ =-125⇔x+1=5⇔x=4
