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a,x.3\(\dfrac{1}{4}\)+(\(\dfrac{-7}{6}\)).x-1\(\dfrac{2}{3}\)=\(\dfrac{5}{12}\)
\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x-\(\dfrac{5}{3}\)=\(\dfrac{5}{12}\)
\(\Leftrightarrow\)\(\dfrac{13}{4}\)x+(\(\dfrac{-7}{6}\))x=\(\dfrac{25}{12}\)
\(\Leftrightarrow\)\(\dfrac{25}{12}\)x=\(\dfrac{25}{12}\)
\(\Leftrightarrow\)x=1
Vậy x=1
b,\(\dfrac{2}{5}\)+\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)
\(\Leftrightarrow\)\(\dfrac{3}{5}\).(3x-3,7)=\(\dfrac{-53}{10}\)-\(\dfrac{2}{5}\)
\(\Leftrightarrow\)\(\dfrac{9}{5}\)x-\(\dfrac{111}{50}\)=\(\dfrac{-57}{10}\)
\(\Leftrightarrow\)\(\dfrac{9}{5}\)x=\(\dfrac{-87}{25}\)
\(\Leftrightarrow\)x=\(\dfrac{-29}{15}\)
Vậy x=\(\dfrac{-29}{15}\)
c,\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)+\(\dfrac{5}{9}\)=\(\dfrac{23}{27}\)
\(\Leftrightarrow\)\(\dfrac{7}{9}\):(2+\(\dfrac{3}{4}\)x)=\(\dfrac{8}{27}\)
\(\Leftrightarrow\)\(\dfrac{7}{18}\)+\(\dfrac{28}{27}\)x=\(\dfrac{8}{27}\)
\(\Leftrightarrow\)\(\dfrac{28}{27}\)x=\(\dfrac{-5}{54}\)
\(\Leftrightarrow\)x=\(\dfrac{-5}{56}\)
Vậy x=\(\dfrac{-5}{56}\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)
\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\).
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)
\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)
\(3x-3\cdot7=-\dfrac{19}{2}\)
\(3x-21=-\dfrac{19}{2}\)
\(3x=-\dfrac{19}{2}+21\)
\(3x=\dfrac{23}{2}\)
\(x=\)\(\dfrac{23}{2}:3\)
\(x=\dfrac{23}{6}\)
Vậy \(x=\dfrac{23}{6}\).
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)
\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)
\(\dfrac{3}{4x}=-\dfrac{65}{22}\)
\(4x=\dfrac{3\cdot22}{-65}\)
\(4x=-\dfrac{66}{65}\)
\(x=-\dfrac{66}{65}:4\)
\(x=-\dfrac{33}{130}\)
Vậy \(x=-\dfrac{33}{130}\).
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)
\(\left|x\right|=\dfrac{29}{12}\)
\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)
Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).
\(a,250:\left(10-x\right)=25\\ \Rightarrow10-x=10\\ \Rightarrow x=0.\\b,3x-2018:2=23\\ \Rightarrow3x-1009=23\\ \Rightarrow3x=1032\\ \Rightarrow x=344. \)
a) \(250:\left(10-x\right)=25\)
\(10-x=250:25\)
\(10-x=10\)
\(x=10-10\)
\(x=0\)
b) \(3x-2018:2=23\)
\(3x-1009=23\)
\(3x=23+1009\)
\(3x=1032\)
\(x=1032:3\)
\(x=344\)
a) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)
=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-57}{10}\)
=> \(\left(3x-3,7\right)=\frac{-19}{5}\)
=>\(3x=\frac{-19}{5}+\frac{37}{10}\)
=>\(3x=\frac{-1}{10}\)
=>\(x=\frac{-1}{30}\)
Ko biết đúng hay sai!!?
\(a,250:\left(10-x\right)=25\\ \Rightarrow10-x=250:25\\ \Rightarrow10-x=10\\ \Rightarrow x=10-10=0\\ b,3x-2018:2=23\\ \Rightarrow3x-1009=23\\ \Rightarrow3x=23+1009\\ \Rightarrow3x=1032\\ \Rightarrow x=1032:3=344\\ c,\left(9x-21\right):3=2\\ \Rightarrow9x-21=2\times3\\ \Rightarrow9x-21=6\\ \Rightarrow9x=21+6\\ \Rightarrow9x=27\\ \Rightarrow x=27:9=3\)
\(d,53\left(9-x\right)=53\\ \Rightarrow9-x=53:53\\ \Rightarrow9-x=1\\ \Rightarrow x=9-1=8\)