|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)
\(\dfrac{2}{3}\) - |\(x+2\)| = 2
|\(x+2\)| = \(\dfrac{2}{3}\) - 2
|\(x+2\)| = - \(\dfrac{4}{3}\)
|\(x+2\)| ≥ 0 ∀ \(x\)
Vậy \(x\in\) \(\varnothing\)
\(\dfrac{5}{6}\).|3 - \(x\)| + 4 = 2
\(\dfrac{5}{6}\).| 3 - \(x\)| = 2 - 4
\(\dfrac{5}{6}\).|3 - \(x\)| = -2
|3- \(x\)| ≥ 0
\(\dfrac{5}{6}\).|3 - \(x\)| ≥ 0 ∀ \(x\)
vậy \(x\in\) \(\varnothing\)