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\(a.\frac{x}{-15}=\frac{60}{2}\)
\(\Rightarrow2x=60\times\left(-15\right)\)
\(\Rightarrow2x=-900\)
\(\Rightarrow x=-900\div2\)
\(\Rightarrow x=-450\)
a) \(\frac{18}{45}-\frac{4}{12}-\frac{6}{9}-\frac{21}{35}=\frac{2}{5}-\frac{1}{3}-\frac{2}{3}-\frac{3}{5}\)
\(=\left(\frac{2}{5}-\frac{3}{5}\right)-\left(\frac{1}{3}+\frac{2}{3}\right)\)
\(=\frac{-1}{3}-1\)
\(=\frac{-4}{3}\)
b) \(\frac{4}{3}+\frac{3}{5}+\frac{7}{3}+\frac{2}{5}+\frac{1}{3}=\left(\frac{4}{3}+\frac{7}{3}+\frac{1}{3}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=4+1=5\)
c) \(\frac{1}{3}.\frac{4}{5}.\frac{1}{3}.\frac{6}{5}=\frac{8}{75}\)
d) \(\frac{6}{7}.\frac{8}{13}+\frac{6}{7}.\frac{9}{13}-\frac{3}{13}.\frac{6}{7}\)
\(=\frac{6}{7}.\left(\frac{8}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{6}{7}.\frac{14}{13}\)
\(=\frac{12}{13}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a) = -120
b) = 3 x (-4)
= -12
c) = -8 x 2
= -16
d) = -18 : (-6)
= 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/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)
=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)
=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)
=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)
13/3: x/4=6:0,3
=>13/3: x/4 =20
=>x/4=(13/3):20
=>x/4=260/3
=>x= (260/3 ) x4
=> x= 1040/3
13/3: x/4=6:0,3
=>13/3: x/4 =20
=>x/4=(13/3):20
=>x/4=260/3
=>x= (260/3 ) x4
=> x= 1040/3