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a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
a )
\(\frac{-4}{9}.\frac{1}{3}-\frac{4}{9}.\frac{5}{6}+\frac{3}{7}.\frac{4}{9}\)
\(=\frac{4}{9}.\left(-\frac{1}{3}-\frac{5}{6}+\frac{3}{7}\right)\)
\(=\frac{4}{9}.\left(-\frac{14}{42}-\frac{35}{42}+\frac{18}{42}\right)\)
\(=\frac{4}{9}.\frac{-31}{42}\)
\(=-\frac{62}{189}\)
b )
\(\frac{2}{3}:\frac{3}{7}-\frac{2}{3}:\frac{4}{3}+\frac{2}{3}:\frac{1}{21}\)
\(=\frac{2}{3}.\frac{7}{3}-\frac{2}{3}.\frac{3}{4}+\frac{2}{3}.21\)
\(=\frac{14}{9}-\frac{1}{2}+14\)
\(=\frac{28}{18}-\frac{9}{18}+14\)
\(=\frac{19}{18}+14\)
\(=1+14+\frac{1}{18}\)
\(=15\frac{1}{18}\)
c )
\(\left(5\frac{1}{3}+3\frac{2}{3}\right)-4\frac{1}{3}\)
\(=\left(5+3-4\right)+\left(\frac{1}{3}+\frac{2}{3}-\frac{1}{3}\right)\)
\(=4\frac{2}{3}\)
\(=\frac{14}{3}\)
a) \(-\frac{4}{9}\cdot\frac{1}{3}-\frac{4}{9}\cdot\frac{5}{6}+\frac{3}{7}\cdot\frac{4}{9}\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{1}{3}+\left(-\frac{4}{9}\right)\cdot\frac{5}{6}-\left(-\frac{4}{9}\right)\cdot\frac{3}{7}\)
\(=\left(-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{5}{6}-\frac{3}{7}\right)\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{31}{42}=-\frac{62}{189}\)
\(\Rightarrow\left(\frac{3x-2}{6}\right):\left(\frac{9-1-6}{6}\right)=\frac{1}{9}\)
\(\Rightarrow\left(\frac{3x-1}{6}\right):\frac{1}{3}=\frac{1}{9}\)
\(\Rightarrow\frac{3x-1}{2}=\frac{1}{9}\)
\(\Rightarrow27x-9=2\)
\(\Rightarrow27x=11\)
\(\Rightarrow x=\frac{11}{27}\)
\(\Rightarrow\left(\frac{x}{2}-\frac{1}{3}\right):\left(\frac{8}{6}-\frac{6}{6}\right)=\frac{1}{9}\)
\(\Rightarrow\left(\frac{x}{2}-\frac{1}{3}\right)=\frac{1}{9}\times\frac{1}{3}\)
\(\Rightarrow\frac{x}{2}=\frac{1}{27}+\frac{1}{3}\)
\(\Rightarrow\frac{x}{2}=\frac{10}{27}\)
\(\Rightarrow10\times2=27x=20\)
\(\Rightarrow x=20:27=\frac{20}{27}\)
Ta có :
\(S=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)
\(\Leftrightarrow\)\(3S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Leftrightarrow\)\(3S-S=\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\right)\)
\(\Leftrightarrow\)\(2S=\frac{1}{3}-\frac{1}{3^9}\)
\(\Leftrightarrow\)\(2S=\frac{3^8-1}{3^9}\)
\(\Leftrightarrow\)\(S=\frac{3^8-1}{2.3^9}\)
Ở đây mk chỉ ghi \(...\) cho nhanh nếu bạn làm vào vở thì ghi đầy đủ ra nhé
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(A=\frac{1}{3}+\frac{1}{3+6}+\frac{1}{3+6+9}+...+\frac{1}{3+6+9+...+2016}\)
<=> \(3A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+672}\)
Áp dụng công thức: 1+2+3+...+n=n(n+1):2
=> \(3A=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{672.673}{2}}\)
=> \(3A=1+\frac{1}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{672.673}\right)=1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{672}-\frac{1}{673}\right)\)
=> \(3A=1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{673}\right)=1+\frac{671}{2692}=\frac{3363}{2692}\)
=> A = 1121/2692