Đáp án + Giải thích các bước giải:
`|(1)/(2)x+(3)/(5)|=(1)/(2)`
`->` \(\left[ \begin{array}{l}\dfrac{1}{2}x+\dfrac{3}{5}=\dfrac{1}{2}\\\dfrac{1}{2}x+\dfrac{3}{5}=-\dfrac{1}{2}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}\dfrac{1}{2}x=-\dfrac{1}{10}\\\dfrac{1}{2}x=-\dfrac{11}{10}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-\dfrac{1}{5}\\x=-\dfrac{11}{5}\end{array} \right.\)
Vậy `x∈{-(1)/(5);-(11)/(5)}`
``
`|2x-1|-(1)/(2)=(2)/(3)`
`->|2x-1|=(7)/(6)`
`->` \(\left[ \begin{array}{l}2x-1=\dfrac{7}{6}\\2x-1=-\dfrac{7}{6}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}2x=\dfrac{13}{6}\\2x=-\dfrac{1}{6}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=\dfrac{13}{12}\\x=-\dfrac{1}{12}\end{array} \right.\)
Vậy `x∈{(13)/(12);-(1)/(12)}`
``
`|x-3.(1)/(2)|-(1)/(2)=5`
`->|x-(3)/(2)|=(11)/(2)`
`->` \(\left[ \begin{array}{l}x-\dfrac{3}{2}=\dfrac{11}{2}\\x-\dfrac{3}{2}=-\dfrac{11}{2}\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=7\\x=-4\end{array} \right.\)
Vậy `x∈{7;-4}`
