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\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
a, \(\dfrac{x-1}{21}\) = \(\dfrac{3}{x+1}\)
( x-1)(x+1) = 21.3
x2 + x - x -1 = 63
x2 = 63 + 1
x2 = 64
x = + - 8
b, 2\(\dfrac{1}{2}\)x + x = 2\(\dfrac{1}{17}\)
x( \(\dfrac{5}{2}\) + 1) = \(\dfrac{35}{17}\)
x = \(\dfrac{35}{17}\) : ( \(\dfrac{5}{2}\)+1)
x = \(\dfrac{35}{17}\) x \(\dfrac{2}{7}\)
x = \(\dfrac{10}{17}\)
c, (x + \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\) ) : ( 2 + \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)): \(\dfrac{23}{12}\) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)) = \(\dfrac{7}{46}\) x \(\dfrac{23}{12}\)
x - \(\dfrac{5}{12}\) = \(\dfrac{7}{12}\)
x = \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)
x = 1
d, 2\(\dfrac{1}{3}\)x - 1\(\dfrac{3}{4}\)x + \(2\dfrac{2}{3}\) = 3\(\dfrac{3}{5}\)
x( \(\dfrac{7}{3}\) - \(\dfrac{7}{4}\)) + \(\dfrac{8}{3}\) = \(\dfrac{18}{5}\)
x\(\dfrac{7}{12}\) = \(\dfrac{18}{5}\) - \(\dfrac{8}{3}\)
x\(\dfrac{7}{12}\) = \(\dfrac{14}{15}\)
x = \(\dfrac{14}{15}\) : \(\dfrac{7}{12}\)
x = \(\dfrac{8}{5}\)
a: x=4/27-2/3=4/27-18/27=-14/27
b: =>3/4x-1/4x=1/6+7/3
=>1/2x=1/6+14/6=5/2
hay x=5
c: =>13/10x=7/2+5/2=6
=>x=13/10:6=13/60
d: (3x+2)(-2/5x-7)=0
=>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-35/2
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
a) \(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\Leftrightarrow\dfrac{3}{7}x-\dfrac{7}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{1}{2}+\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{17}{6}\)
\(\Leftrightarrow x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(\Leftrightarrow x=\dfrac{119}{18}\)
b) \(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{7}-\dfrac{2}{3}:x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{4}{7}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{56}{-57}\)
c) \(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\Leftrightarrow\left(\dfrac{2}{3}x+\dfrac{9}{4}\right):\dfrac{16}{6}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=\dfrac{3}{4}.\dfrac{16}{6}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=2\)
\(\Leftrightarrow\dfrac{2}{3}x=2-\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{-1}{4}\)
\(\Leftrightarrow x=\dfrac{-1}{4}:\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{-3}{8}\)
d) \(\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}-\dfrac{2}{3}=1\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=1+\dfrac{2}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{3}.\dfrac{1}{4}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4}{5}-\dfrac{2}{3}x=\dfrac{5}{12}\\\dfrac{4}{5}-\dfrac{2}{3}x=-\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{5}{12}\\\dfrac{2}{3}x=\dfrac{4}{5}-\left(-\dfrac{5}{12}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{23}{60}\\\dfrac{2}{3}x=\dfrac{73}{60}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{60}:\dfrac{2}{3}\\x=\dfrac{73}{60}:\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{73}{20}\end{matrix}\right.\)
a)\(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\dfrac{3}{7}x=0,5+2\dfrac{1}{3}\)
\(\dfrac{3}{7}x=\dfrac{17}{6}\)
\(x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(x=\dfrac{119}{18}\)
b)\(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=\dfrac{4}{7}-1\dfrac{1}{4}\)
\(x=\dfrac{2}{3}:\left(-\dfrac{19}{28}\right)\)
\(x=-\dfrac{56}{57}\)
c)\(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\dfrac{2}{3}x+2\dfrac{1}{4}=0,75:3\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{15}{64}-2\dfrac{1}{4}\)
\(x=-\dfrac{129}{64}:\dfrac{2}{3}\)
\(x=-\dfrac{387}{128}\)