4\(\frac{1}{3}\):x/4=6:0,3
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\(\frac{13}{3}:\frac{x}{4}=6:0,3\)
\(\frac{13}{3}:\frac{x}{4}=20\)
\(\frac{x}{4}=\frac{13}{3}:20\)
\(\frac{x}{4}=\frac{13}{3.20}\)
\(\frac{x}{4}=\frac{13}{60}\)
\(60x=13.4\)
\(60x=52\)
\(x=\frac{52}{60}\)
\(x=\frac{13}{15}\)
Vậy \(x=\frac{13}{15}\)
\(\frac{13}{3}:\frac{x}{4}=6:0,3\)
\(\frac{13}{3}:\frac{x}{4}=20\)
\(\frac{x}{4}=\frac{13}{3}:20\)
\(\frac{x}{4}=\frac{13}{60}\)
x.60:=4.13
x.60=52
x=52:60
\(x=\frac{13}{15}\)
vậy \(x=\frac{13}{15}\)
\(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
\(\Rightarrow\frac{13}{3}.\frac{4}{x}=20\)
\(\Rightarrow\frac{4}{x}=\frac{60}{13}\)
\(\Leftrightarrow60x=13.4=52\)
\(\Leftrightarrow x=\frac{13}{15}\)
\(1\frac{2}{3}:\frac{x}{4}=6:0,3\)
\(\frac{5}{3}:\frac{x}{4}=20\)
\(\Rightarrow\frac{x}{4}=\frac{5}{3}:20\)
\(\Rightarrow\frac{x}{4}=\frac{1}{12}\Rightarrow x=\frac{4}{12}=\frac{1}{3}\)
a) Ta có: \(\left(x-2\right)^3+\frac{8}{27}=0\)
\(\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow x-2=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+2=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
b) Ta có: \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
\(\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20\)
\(\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}\)
hay \(x=\frac{13\cdot4}{60}=\frac{13}{15}\)
Vậy: \(x=\frac{13}{15}\)
c) Ta có: \(\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
\(\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}\)
\(\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}\)
\(\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16\)
\(\Leftrightarrow3x=-160\)
hay \(x=\frac{-160}{3}\)
Vậy: \(x=\frac{-160}{3}\)
d) Ta có: \(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{4}{3}\right\}\)
a/ (x - 2)3 + \(\frac{8}{27}\) = 0
=> (x - 2)3 = 0 - \(\frac{8}{27}\) = \(\frac{-8}{27}\)
=> x - 2 = \(-\frac{2}{3}\)
=> x = \(-\frac{2}{3}+2=\frac{4}{3}\)
b/ \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
=> \(4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20\)
=> \(\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}\)
=> \(x=\frac{13}{60}.4=\frac{13}{15}\)
c/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)
=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)
=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)
=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)
\(1\frac{2}{3}:\frac{x}{4}=6:0,3\)
\(\frac{5}{3}:\frac{x}{4}=20\)
\(\frac{x}{4}=\frac{5}{3}:20\)
\(\frac{x}{4}=\frac{1}{12}\)
\(\Rightarrow\frac{3x}{12}=\frac{1}{12}\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)
Vậy \(x=\frac{1}{3}\)
a) \(\frac{-3}{7}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{3}{7}\)
\(\Leftrightarrow x=\frac{16}{21}\)
b) \(\frac{-5}{8}+x=\frac{-2^2}{3}\)
\(\Leftrightarrow x=\frac{-4}{3}+\frac{5}{8}\)
\(\Leftrightarrow x=-\frac{17}{24}\)
c) \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
\(\Leftrightarrow\frac{8}{3}:\frac{x}{4}=20\)
\(\Leftrightarrow\frac{32}{3x}=20\)
\(\Leftrightarrow3x=\frac{8}{5}\)
\(\Leftrightarrow x=\frac{8}{15}\)
Ta có: \(\frac{13}{3}\div\frac{x}{4}=6\div0,3\Rightarrow\frac{13}{3}\div\frac{x}{4}=20\Rightarrow\frac{x}{4}=\frac{13}{3}\div20\Rightarrow\frac{x}{4}=\frac{13}{60}\Rightarrow x=\frac{13}{15}\)