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Bài 1:
\(=\dfrac{-3-39}{42}+\dfrac{-6-11}{17}-\dfrac{1}{6}=-\dfrac{119}{48}\)
Bài 2:
=>x:5=-13/20
hay x=-65/20=-13/4
Bài 1:
\(=\dfrac{-3-39}{32}+\dfrac{-6-11}{17}+\dfrac{-1}{6}=-\dfrac{21}{16}+\dfrac{-1}{6}-1=-\dfrac{119}{48}\)
Bài 2:
\(\Leftrightarrow x:5=-\dfrac{13}{20}\)
hay x=-13/4
\(\frac{5}{6}+\frac{\left(-19\right)}{30}=\frac{x}{5}\)
\(\frac{25}{30}+\frac{\left(-19\right)}{30}=\frac{x}{5}\)
\(\frac{6}{30}=\frac{x}{5}\)
\(\frac{1}{5}=\frac{x}{5}\)
\(\Rightarrow5x=5\times1\)
\(5x=5\)
\(x=5\div5\)
\(x=1\)
\(\frac{5}{6}+\frac{-19}{30}=\frac{x}{5}\)
\(\Leftrightarrow\frac{x}{5}=\frac{5}{6}+\frac{-19}{30}\)
\(\Leftrightarrow\frac{x}{5}=\frac{25}{30}+\frac{-19}{30}\)
\(\Leftrightarrow\frac{x}{5}=\frac{6}{30}\)
\(\Leftrightarrow\frac{x}{5}=\frac{1}{5}\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
Ta có: x(1+30%)=-1,31
⇒ \(\dfrac{13}{10}\)x =-1,31
⇒ x =-1,31 : \(\dfrac{13}{10}\)
⇒ x = -\(\dfrac{131}{130}\)
x + 30% x = -1,31
\(x+\dfrac{30}{100}x=-1,31\)
x+0,3x=-1,31
x.(0,3+1)=-1,31
x.1,3=-1,31
\(x=\dfrac{-1,31}{1,3}\)
\(x=-\dfrac{131}{130}\)
\(x+\left(x+1\right)+\left(x+2\right)+...+19+20=20\)
\(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)+...+19=0\)
\(\Leftrightarrow\frac{x\left(19+x\right)}{2}=0\)
\(\Leftrightarrow x\left(x+19\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+19=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-19\end{cases}}\)
Mà \(x\ne0\)nên \(x=-19\)
\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)
\(x=\dfrac{-42}{20}+\dfrac{5}{20}=\dfrac{-37}{20}\)
\(\dfrac{x}{5}=\dfrac{5}{6}-\dfrac{19}{30}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{6}{30}=\dfrac{1}{5}\)
\(\Leftrightarrow x=1\)
Ta có : \(\dfrac{x}{5}=\dfrac{5}{6}+\left(-\dfrac{19}{30}\right)=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{1}{5}\)
\(\Leftrightarrow5x=5\)
\(\Leftrightarrow x=1\)
Vậy ...