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\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
a.\(x-\dfrac{2}{3}=\dfrac{8}{7}\)
\(x=\dfrac{8}{7}+\dfrac{2}{3}\)
x=\(\dfrac{38}{21}\)
b.\(\left(x+\dfrac{1}{3}\right)=\dfrac{4}{25}
\)
x=\(\dfrac{4}{25}-\dfrac{1}{3}\)
x=\(-\dfrac{13}{75}\)
c.\(-\dfrac{2}{3}:x+\dfrac{5}{8}=-\dfrac{7}{12}\)
\(-\dfrac{2}{3}:x=-\dfrac{29}{24}\)
x=\(\dfrac{16}{29}\)
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
\(-\dfrac{3}{4}\left(\dfrac{8}{9}-x\right)+\dfrac{3}{5}=-\dfrac{2}{3}.\dfrac{1}{2}=-\dfrac{1}{3}\\ \Leftrightarrow-\dfrac{3}{4}\left(\dfrac{8}{9}-x\right)=-\dfrac{1}{3}-\dfrac{3}{5}=-\dfrac{14}{15}\\ \Leftrightarrow\dfrac{8}{9}-x=\dfrac{56}{45}\\ \Leftrightarrow x=\dfrac{8}{9}-\dfrac{56}{45}=-\dfrac{16}{45}\)
Ta có : \(PT\Leftrightarrow-\dfrac{3}{4}\left(\dfrac{8}{9}-x\right)=-\dfrac{2}{3}.\dfrac{1}{2}-\dfrac{3}{5}=-\dfrac{14}{15}\)
\(\Rightarrow\dfrac{8}{9}-x=\dfrac{-\dfrac{14}{15}}{-\dfrac{3}{4}}=\dfrac{56}{45}\)
\(\Rightarrow x=\dfrac{8}{9}-\dfrac{56}{45}=-\dfrac{16}{45}\)
Vậy ...
|x+1|=\(\dfrac{8}{3}\)-\(\dfrac{2}{3}\)
|x+1|=2
TH1: x+1=2 TH2: x+1=-2
x=2-1 x=-2-1
x=1 x=-3
vậy x={1;-3}
\(\dfrac{8}{3}-\left|x+1\right|=\dfrac{2}{3}\)
\(\left|x+1\right|=\dfrac{8}{3}-\dfrac{2}{3}\)
\(\left|x+1\right|=2\)
\(x+1=2\) hoặc \(x+1=-2\)
TH1 : \(x+1=2\)
\(x=2-1\)
\(x=1\)
TH2 : \(x+1=-2\)
\(x=-2-1\)
\(x=-3\)
Vậy x = 1 hoặc x = -3