Tìm x, biết: \(\dfrac{1}{5}-\dfrac{1}{2}:\left(0,5x-1,5\right)=0,35\)
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a: =>2480-1570+(205-x)=1010
=>205-x=100
hay x=105
b: \(\Leftrightarrow x\cdot\dfrac{1}{2}\cdot\dfrac{13}{12}=\dfrac{7}{60}\cdot3\cdot4=\dfrac{7}{5}\)
=>x=168/65
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
a. \(\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\Rightarrow2x=\dfrac{1}{5}\Rightarrow x=\dfrac{1}{5}:2=0,1\)
Vậy \(x=0,1\)
b. \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\Rightarrow-1\dfrac{3}{5}+x=\dfrac{13}{6}\cdot\dfrac{12}{13}=2\Rightarrow x=2+1\dfrac{3}{5}=3,6\)
Vậy \(x=3,6\)
c. \(-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\Rightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\Rightarrow-\dfrac{4}{7}x=-\dfrac{3}{40}-\dfrac{7}{5}=-\dfrac{59}{40}\Rightarrow x=\left(-\dfrac{59}{40}\right):\left(-\dfrac{4}{7}\right)=2,58125\)
Vậy \(x=2,58125\)
a, \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{11}{12}\)
\(\Rightarrow x=11\)
b, \(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{19}{15}x=\dfrac{-3}{5}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{15}\)
\(\Rightarrow x=1\)
c, \(-2^3+0,5x=1,5\)
\(\Rightarrow-8+\dfrac{1}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{2}+8\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{19}{2}\)
\(\Rightarrow x=19\)
1) \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\) 2)\(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\) \(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}\) \(\dfrac{2}{3}-\dfrac{19}{15}x=\dfrac{-3}{5}\) \(\dfrac{x}{12}=\dfrac{11}{12}\) \(\dfrac{19}{15}x=\dfrac{2}{3}-\left(\dfrac{-3}{5}\right)\) => \(x=11\) \(\dfrac{19}{15}x=\dfrac{19}{15}\) => \(x=1\) 3) -23 + 0,5x = 1,5 -8 + 0,5x = 1,5 0,5x = 1,5 - (-8) 0,5x = 9,5 x = 9,5 : 0,5 x = 19
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
a.\(\dfrac{-4}{5}-\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{2}{7}\)
\(\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{-4}{5}-\dfrac{2}{7}=\dfrac{-38}{35}\)
\(\dfrac{2}{3}x=\dfrac{-38}{35}-1\dfrac{1}{4}\)
\(\dfrac{2}{3}x=\dfrac{-327}{140}\Rightarrow x=\dfrac{-327}{140}:\dfrac{2}{3}=\dfrac{-981}{280}\)
Vậy \(x=\dfrac{-981}{280}\)
b. \(\dfrac{5}{6}+\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}-\dfrac{5}{6}=\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{3}{4}-\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{9}{4}\Rightarrow x=\dfrac{1}{2}:\dfrac{9}{4}=\dfrac{2}{9}\)
Vậy \(x=\dfrac{2}{9}\)
c. \(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right):\dfrac{3}{4}=0,7\)
\(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right)=0,7.\dfrac{3}{4}=\dfrac{21}{40}\)
\(\dfrac{4}{5}x=\dfrac{21}{40}+1\dfrac{1}{3}=\dfrac{223}{120}\)
\(\Rightarrow x=\dfrac{223}{120}:\dfrac{4}{5}=\dfrac{223}{96}\)
Vậy \(x=\dfrac{223}{96}\)
d. \(\dfrac{5}{6}-\dfrac{3}{4}x=1\dfrac{1}{3}+0,5x\)
\(0,5x+\dfrac{3}{4}x=\dfrac{5}{6}-1\dfrac{1}{3}\)
\(\dfrac{5}{4}x=\dfrac{-1}{2}\Rightarrow x=\dfrac{-1}{2}:\dfrac{5}{4}=\dfrac{-2}{5}\)
Vậy \(x=\dfrac{-2}{5}\)
=>1/2:(0,5x-1,5)=0,2-0,35=-0,15=-3/20
=>0,5x-1,5=1/2:(-3/20)=-1/2*20/3=-10/3
=>0,5x=-10/3+3/2=-11/6
=>x=-11/3