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a) \(\dfrac{5}{6}:x=30:3\)
\(\Leftrightarrow\dfrac{5}{6}:x=10\)
\(\Leftrightarrow x=\dfrac{5}{6}:10\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy .......
b) \(x:2,5=0,003:0,75\)
\(\Leftrightarrow x:2,5=0,004\)
\(\Leftrightarrow x=0,004.2,5\)
\(\Leftrightarrow x=0,01\)
Vậy .......
c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)
\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)
\(\Leftrightarrow2x=\dfrac{698}{25}\)
\(\Leftrightarrow x=\dfrac{304}{15}\)
Vậy ...
d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)
\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{8}{15}\)
Vậy ....
e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)
\(\Leftrightarrow0,25x=\dfrac{57}{608}\)
\(\Leftrightarrow x=\dfrac{228}{608}\)
Vậy ...
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)
\(\Leftrightarrow x^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy ...
a. \(A=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
Vậy \(A=\dfrac{3}{11}\)
b. \(B=\dfrac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{2^{12}\left(13+65\right)}{2^{10}\cdot104}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\dfrac{2^{12}\cdot78}{2^{10}\cdot104}+\dfrac{3^{10}\cdot16}{3^9\cdot16}=\dfrac{2^2\cdot3}{1\cdot4}+3=\dfrac{12}{4}+3=3+3=6\)
Vậy \(B=6\)
A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
a) \(\dfrac{27}{2}\cdot7,5+\dfrac{27}{2}\cdot2,5-150\)
\(=\dfrac{27}{2}\cdot\left(7,5+2,5\right)-150\)
\(=\dfrac{27}{2}\cdot10-150\)
\(=135-150\)
\(=-15\)
b) \(3^3\cdot\dfrac{18}{5}-3^3\cdot2\dfrac{2}{5}-3^3\cdot\dfrac{6}{5}\)
\(=3^3\cdot\dfrac{18}{5}-3^3\cdot\dfrac{12}{5}-3^3\cdot\dfrac{6}{5}\)
\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{12}{5}-\dfrac{6}{5}\right)\)
\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{18}{5}\right)\)
\(=3^3\cdot0\)
\(=0\)
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)
= \(\dfrac{3}{11}\)
b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
= \(\dfrac{3}{5}\)
c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)
= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)
= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)
= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)
A= \(\dfrac{11}{9}\).\(\dfrac{-3}{2}\)- \(\dfrac{11}{9}\).\(\dfrac{15}{2}\)+(2021)0
= \(\dfrac{11}{9}\)(\(\dfrac{-3}{2}\)-\(\dfrac{15}{2}\)) + 1
= \(\dfrac{11}{9}\)(-9) + 1
= -11 + 1
= -10
Lần sau bn ko đc phép chuyển \(\left(-2021\right)^0\) sang \(\left(2021\right)^0\) đou nha