Giải thích các bước giải:
4/ $|\dfrac{3}{2}x+\dfrac{1}{2}|=|4x-1|$
⇒ \(\left[ \begin{array}{l}\dfrac{3}{2}x+\dfrac{1}{2}=4x-1\\\dfrac{3}{2}x+\dfrac{1}{2}=-4x+1\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}\dfrac{5}{2}x=\dfrac{3}{2}\\\dfrac{11}{2}x=\dfrac{1}{2}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{1}{11}\end{array} \right.\)
5/ $|\dfrac{7}{5}x+\dfrac{1}{2}|=|\dfrac{4}{3}x-\dfrac{1}{4}|$
⇒ \(\left[ \begin{array}{l}\dfrac{7}{5}x+\dfrac{1}{2}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{1}{2}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}\dfrac{1}{15}x=-\dfrac{3}{4}\\\dfrac{41}{15}x=-\dfrac{1}{4}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-\dfrac{45}{4}\\x=-\dfrac{15}{164}\end{array} \right.\)
6/ $(4x-9)(2,5+\dfrac{-7}{3}x)=0$
⇒ \(\left[ \begin{array}{l}4x-9=0\\2,5-\dfrac{7}{3}x=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\dfrac{9}{4}\\x=\dfrac{15}{14}\end{array} \right.\)
7/ $\dfrac{3}{2}-(x-\dfrac{5}{6})=\dfrac{8}{9}$
⇒ $x-\dfrac{5}{6}=\dfrac{3}{2}-\dfrac{8}{9}$
⇒ $x-\dfrac{5}{6}=\dfrac{11}{18}$
⇒ $x=\dfrac{11}{18}+\dfrac{5}{6}$
⇒ $x=\dfrac{13}{9}$
Chúc bạn học tốt !!!
