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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: =>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 -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
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
\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)
\(x=\dfrac{7}{35}+\dfrac{-15}{35}\)
\(x=-\dfrac{8}{35}\)
\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{2}\)
\(x=\dfrac{4}{7}:\dfrac{3}{2}\)
\(x=\dfrac{4}{7}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{21}\)
\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)
\(x+\dfrac{3}{4}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}-\dfrac{3}{4}\)
\(x=-\dfrac{23}{12}\)
\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{13}{18}\)
\(x=\dfrac{-5}{9}-\dfrac{13}{18}\)
\(x=\dfrac{-10}{18}-\dfrac{13}{18}\)
\(x=-\dfrac{23}{18}\)
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
1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
\(\dfrac{3}{7}-x=\dfrac{1}{4}-\dfrac{-3}{5}\\ \Leftrightarrow\dfrac{3}{7}-x=\dfrac{17}{20}\\ \Rightarrow-x=\dfrac{17}{20}-\dfrac{3}{7}=\dfrac{59}{140}\\ \Rightarrow x=-\dfrac{59}{140}\)
\(\left|x-\dfrac{3}{5}\right|-\dfrac{3}{4}=-\dfrac{1}{2}\\\Leftrightarrow \left|x-\dfrac{3}{5}\right|=-\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{1}{4}\\x-\dfrac{3}{5}=-\dfrac{1}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{7}{20}\end{matrix}\right.\)
\(\dfrac{3}{7}-x=\dfrac{1}{4}-\dfrac{-3}{5}\\ \dfrac{3}{7}-x=\dfrac{17}{20}\\ x=\dfrac{3}{7}-\dfrac{17}{20}\\ x=-\dfrac{59}{140}\)
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\(\left|x-\dfrac{3}{5}\right|-\dfrac{3}{4}=-\dfrac{1}{2}\\ \left|x-\dfrac{3}{5}\right|=-\dfrac{1}{2}+\dfrac{3}{4}\\ \left|x-\dfrac{3}{5}\right|=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{1}{4}\\x-\dfrac{3}{5}=-\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=-\dfrac{7}{20}\end{matrix}\right.\)