b)
\(\dfrac{1}{2}\left(\dfrac{4}{5}-\dfrac{3}{2}\right)+x=5\left(x-\dfrac{1}{3}\right)\)
=> \(-\dfrac{7}{20}+x=5x-\dfrac{5}{3}\)
=> \(\dfrac{79}{60}=4x\)
=> \(\dfrac{79}{240}=x\)
Vậy \(\dfrac{79}{240}=x\)
c)
\(2\left(x-5\right)-3\left(x+7\right)=14\)
=> \(2x-10-3x-21=14\)
=> \(-x-31=14\)
=> \(-x=45\)
=> x = -45
Vậy x là -45
d)
-8 . |x - 3| = -32
=> |x - 3| = 4
=> \(\left\{{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)
=> x = 7 và x = -1
Vậy x \(\in\left\{-1;7\right\}\)
e)
-90 : |x + \(\dfrac{3}{4}\)| = -45
=> |x + \(\dfrac{3}{4}\)| = -90 : -45 = 2
=> \(\left\{{}\begin{matrix}x+\dfrac{3}{4}=2\\x+\dfrac{3}{4}=-2\end{matrix}\right.\)
=> x = \(\dfrac{5}{4}\) và x = \(-\dfrac{11}{4}\)
Vậy x \(\in\left\{-\dfrac{11}{4};\dfrac{5}{4}\right\}\)
Giải:
b, \(\dfrac{1}{2}.\left(\dfrac{4}{5}-\dfrac{3}{2}\right)+x=5.\left(x-\dfrac{1}{3}\right)\)
\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{8}{10}-\dfrac{8}{10}\right)+x=5x-\dfrac{5}{3}\)
\(\Rightarrow\dfrac{1}{2}.0+x=5x-\dfrac{5}{3}\)
\(\Rightarrow x=5x-\dfrac{5}{3}\)
\(\Leftrightarrow5x-\dfrac{5}{3}=x\)
\(\Rightarrow5x-x=\dfrac{5}{3}\)
\(\Rightarrow4x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{12}\)
c, 2.(x - 5) - 3.(x + 7) = 14
\(\Rightarrow\left(2x-10\right)-\left(3x+21\right)=14\)
\(\Rightarrow\) 2x - 10 - 3x - 21 = 14
\(\Rightarrow-x-31=14\)
\(\Rightarrow-x=45\)
\(\Rightarrow x=-45\)
d, \(-8.\left|x-3\right|=-32\)
\(\Rightarrow\left|x-3\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
e, \(-90:\left|x+\dfrac{3}{4}\right|=-45\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=2\\x+\dfrac{3}{4}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
