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a: =>|x+3/4|=2+1/5=11/5
=>x+3/4=11/5 hoặc x+3/4=-11/5
=>x=29/20 hoặc x=-59/20
b: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
c: =>|2x-1/3|=1/6
=>2x-1/3=1/6 hoặc 2x-1/3=-1/6
=>2x=1/2 hoặc 2x=1/6
=>x=1/4 hoặc x=1/12
e: =>x+2/3=0 hoặc -2x-3/5=0
=>x=-2/3 hoặc x=-3/10
a)<=>10+15(3x-3,7)=-5,3
<=>10+45x-55,5=-5,3
<=>45x-45,5=-5,3
<=>45x=40,2
<=>x=\(\frac{40,2}{45}\)
b)
a) \(x=-\frac{1}{2}\)
b) \(x=-\frac{29}{15}\)
c) \(x=\frac{5}{6}\)
**** cho mình nhé
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,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=\(\dfrac{-5}{56}\)
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