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Tìm x
a) | x - \(\dfrac{3}{2}\) | = \(\left(\dfrac{12}{13}+7\right)\) + \(\left(-8+\dfrac{1}{13}\right)\)
| x - \(\dfrac{3}{2}\) | = \(\dfrac{12}{13}+7-8+\dfrac{1}{13}\)
| x - \(\dfrac{3}{2}\) | = 1 + 7 - 8
=> | x - \(\dfrac{3}{2}\) | = 0
=> x - \(\dfrac{3}{2}\) = 0
=> x = \(\dfrac{3}{2}\)
b) ( \(\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) ) . \(\dfrac{1}{30}\) - | 3 . x - 3 | = 2
=> \(\dfrac{27}{4}\) . \(\dfrac{1}{30}\) - | 3x - 3 | = 2
=> \(\dfrac{9}{40}\) - | 3x - 3 | = 2
=> | 3x - 3 | = \(\dfrac{9}{40}\) - 2
=> | 3x - 3 | = \(\dfrac{-71}{40}\)
Th1 :
3x - 3 = \(\dfrac{-71}{40}\)
=> 3x = \(\dfrac{-71}{40}\) + 3
=> 3x = \(\dfrac{49}{40}\)
=> x = \(\dfrac{49}{40}\) : 3
=> x = \(\dfrac{49}{120}\)
TH2 :
3x - 3 = \(\dfrac{71}{40}\)
=> 3x = \(\dfrac{191}{40}\)
=> x = \(\dfrac{191}{120}\)
Vậy x = \(\dfrac{49}{120}\) hoặc \(\dfrac{191}{120}\)
a: =>|x-3/2|=12/13+1/13+7-8=0
=>x-3/2=0
hay x=3/2
b: \(\Leftrightarrow\dfrac{45-12-10}{60}\cdot\dfrac{1}{30}-\left|3x-3\right|=2\)
\(\Leftrightarrow\left|3x-3\right|=\dfrac{23}{60}\cdot\dfrac{1}{30}-2=-\dfrac{1777}{1800}\)(vô lý)