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)

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