\(-1\dfrac{1}{3}=?\)
K
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19 tháng 11 2017
8,A=\(\dfrac{9}{10}-\left(\dfrac{1}{10\times9}+\dfrac{1}{9\times8}+\dfrac{1}{8\times7}+...+\dfrac{1}{2\times1}\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+...+\dfrac{1}{2}-1\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-1\right)\)
=\(\dfrac{9}{10}-\dfrac{\left(-9\right)}{10}\)
=\(\dfrac{9}{5}\)
H
11 tháng 9 2023
\(a,1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\\ =1+\dfrac{1}{1+\dfrac{1}{\dfrac{3}{2}}}\\ =1+\dfrac{1}{1+\dfrac{2}{3}}\\ =1+\dfrac{1}{\dfrac{5}{3}}\\ =1+\dfrac{3}{5}\\ =\dfrac{8}{5}\)
\(b,1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{9}}}\\ =1-\dfrac{1}{1-\dfrac{1}{\dfrac{8}{9}}}\\ =1-\dfrac{1}{1-\dfrac{9}{8}}\\ =1-\dfrac{1}{-\dfrac{1}{8}}\\=1-\left(-8\right)\\ =9\)
\(c,-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{3}}}}\\ =-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{\dfrac{4}{3}}}}\\ =-3+\dfrac{1}{1+\dfrac{1}{3+\dfrac{3}{4}}}\\ =-3+\dfrac{1}{1+\dfrac{1}{\dfrac{15}{4}}}\\ =-3+\dfrac{1}{1+\dfrac{4}{15}}\\ =-3+\dfrac{1}{\dfrac{19}{15}}\\ =-3+\dfrac{15}{19}\\ =-\dfrac{57}{19}+\dfrac{15}{19}\\ =-\dfrac{42}{19}\)
4 tháng 9 2021
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
10 tháng 9 2021
\(=\dfrac{3^5}{3}+\dfrac{3^5}{3^2}+\dfrac{3^5}{3^3}+\dfrac{3^5}{3^4}+\dfrac{3^9}{3^5}+\dfrac{3^9}{3^6}+\dfrac{3^9}{3^7}+\dfrac{3^9}{3^8}\\ =2\left(3^4+3^3+3^2+3\right)\)
MH
10 tháng 9 2021
\(=\left(\dfrac{3^5}{3}+\dfrac{3^5}{3^2}+\dfrac{3^5}{3^3}+\dfrac{3^5}{3^4}\right)+\left(\dfrac{3^9}{3^5}+\dfrac{3^9}{3^6}+\dfrac{3^9}{3^5}+\dfrac{3^9}{3^8}\right)\)
\(=\left(3^4+3^3+3^2+3\right)+\left(3^4+3^3+3^2+3\right)\)
\(=2\left(3^4+3^3+3^2+3\right)\)
\(=2\left(81+27+9+3\right)\)
\(=2\left(120\right)\)
\(=240\)
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