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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)
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}\)
⇒ \(x=\dfrac{21}{52}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)
\(\dfrac{x}{3}=\dfrac{11}{21}\)
⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)
⇒ \(x=\dfrac{11}{7}\)
c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)
\(\dfrac{-17}{7}< x< -1\)
⇒ \(-17< x< -7\)
⇒ \(x\in\left\{-16;-15,....;-6\right\}\)
d) \(\dfrac{1}{6}+\dfrac{2}{5}\)
\(=\dfrac{5}{30}+\dfrac{12}{30}\)
\(=\dfrac{17}{30}\)
e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)
\(=\dfrac{12}{20}+\dfrac{-35}{20}\)
\(=\dfrac{-23}{20}\)
f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)
\(=\dfrac{4}{13}+\dfrac{-2}{5}\)
\(=\dfrac{20}{65}+\dfrac{-26}{65}\)
\(=\dfrac{-6}{65}\)
g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)
\(=\dfrac{-6}{58}+\dfrac{16}{58}\)
\(=\dfrac{10}{58}\)
h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)
\(=\dfrac{1}{5}+\dfrac{-4}{5}\)
\(=\dfrac{-3}{5}\)
j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)
\(=\dfrac{-2}{9}+\dfrac{5}{9}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
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 \(\dfrac{191}{120}\)