Giải thích các bước giải:
a/ $|2x+5|=4$
⇒ \(\left[ \begin{array}{l}2x+5=4\\2x+5=-4\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=-\dfrac{9}{2}\end{array} \right.\)
b/ `|x+\dfrac{1}{4}|-\dfrac{3}{4}=5%`
⇒ $|x+\dfrac{1}{4}|-\dfrac{3}{4}=\dfrac{1}{20}$
⇒ $|x+\dfrac{1}{4}|=\dfrac{1}{20}+\dfrac{3}{4}$
⇒ $|x+\dfrac{1}{4}|=\dfrac{4}{5}$
⇒ \(\left[ \begin{array}{l}x+\dfrac{1}{4}=\dfrac{4}{5}\\x+\dfrac{1}{4}=-\dfrac{4}{5}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=\dfrac{11}{20}\\x=-\dfrac{21}{20}\end{array} \right.\)
c/ $|\dfrac{7}{5}x+\dfrac{2}{3}|=|\dfrac{4}{3}x-\dfrac{1}{4}|$
⇒ \(\left[ \begin{array}{l}\dfrac{7}{5}x+\dfrac{2}{3}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{2}{3}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}\dfrac{1}{15}x=-\dfrac{11}{12}\\\dfrac{41}{15}x=-\dfrac{5}{12}\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=-\dfrac{55}{4}\\x=-\dfrac{25}{164}\end{array} \right.\)
d/ $|5-\dfrac{3}{4}x|+|\dfrac{2}{7}y-3|=0$
Vì $|5-\dfrac{3}{4}x| \geq 0$ và $|\dfrac{2}{7}y-3| \geq 0$
nên $\left\{\begin{matrix}5-\dfrac{3}{4}x=0 & \\\dfrac{2}{7}y-3=0 & \end{matrix}\right.$
⇒ $\left\{\begin{matrix}x=\dfrac{20}{3} & \\y=\dfrac{21}{2} & \end{matrix}\right.$
Chúc bạn học tốt !!!
