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a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(-\dfrac{5}{6}x=\dfrac{5}{12}\)
\(x=-\dfrac{1}{2}\)
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)
\(3x-3.7=-\dfrac{19}{2}\)
\(3x=-5.8\)
\(x=-\dfrac{29}{15}\)
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
\(\dfrac{3}{4}x=\dfrac{5}{8}\)
\(x=\dfrac{5}{6}\)
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=-\dfrac{3}{20}\)
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
\(x:4\dfrac{1}{3}=2,5\\ x:\dfrac{13}{3}=\dfrac{5}{2}\\ x=\dfrac{5}{2}.\dfrac{13}{3}\\ x=\dfrac{65}{6}=10\dfrac{5}{6}\)
b) \(\dfrac{4}{5}-\dfrac{3}{4}:x=0,3\)
\(\Rightarrow0,8-0,75:x=0,3\)
\(\Rightarrow0,75:x=0,5\)
\(\Rightarrow x=1,5\)
c) \(\dfrac{-3}{2}-\dfrac{1}{4}x=1\dfrac{1}{3}-0,2x\)
\(\Rightarrow\dfrac{-3}{2}-\dfrac{4}{3}=\dfrac{1}{4}x-\dfrac{1}{5}x\)
\(\Rightarrow x=\dfrac{-17}{6}\cdot20\)
\(\Rightarrow x=\dfrac{-170}{3}\)
a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)
\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)
\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)
\(x=-\dfrac{1}{11}\)
Vậy \(x=-\dfrac{1}{11}\).
c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)
\(60\%x=\dfrac{19}{9}\)
\(\dfrac{3}{5}x=\dfrac{19}{9}\)
\(x=\dfrac{19}{9}:\dfrac{3}{5}\)
\(x=\dfrac{95}{27}\)
Vậy \(x=\dfrac{95}{27}\).
d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)
\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)
\(\dfrac{2}{3}-x=\dfrac{3}{20}\)
\(x=\dfrac{2}{3}-\dfrac{3}{20}\)
\(x=\dfrac{31}{60}\)
Vậy \(x=\dfrac{31}{60}\).
e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)
\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)
\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)
\(-2x=-\dfrac{73}{20}\)
\(x=-\dfrac{73}{20}:\left(-2\right)\)
\(x=\dfrac{73}{40}\)
Vậy \(x=\dfrac{73}{40}\).
\(a,\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{3}:\dfrac{2}{3}=1\)\(b,5\dfrac{4}{7}:x=13\Leftrightarrow\dfrac{39}{7}:x=13\Leftrightarrow x=\dfrac{39}{7}:13=\dfrac{3}{7}\)\(c,\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\Leftrightarrow\left(\dfrac{14}{5}x-50\right).\dfrac{3}{2}=51\Leftrightarrow\dfrac{21}{5}x-75=51\Leftrightarrow\dfrac{21}{5}x=51+75=126\Leftrightarrow x=126:\dfrac{21}{5}=30\)
d,\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{-1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
x - 2/3 = 3/5 - 1/4
x - 2/3 = 7/20
x = 7/20 + 2/3
x = 61/60
- 2/3 . x + 1/5 = 1/10
-2/3 . x = 1/10 - 1/5
- 2/3 . x = -1/10
x = -1/10 : -2/3
x = 3/20