( 1+\(\dfrac{2}{3}\) ) . ( 1+\(\dfrac{2}{5}\) ) . (1+\(\dfrac{2}{7}\) ) ....(1+\(\dfrac{2}{2017}\) )
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Tính A =\(\dfrac{3^2-1}{5^2-1}.\dfrac{7^2-1}{^{ }9^2-1}......\dfrac{2015^2-1}{2017^2-1}.\dfrac{2017^2-1}{2019^2-1}\)
\(\frac{3^2-1}{5^2-1}.\frac{7^2-1}{9^2-1}......\frac{2015^2-1}{2017^2-1}.\frac{2017^2-1}{2019^2-1}\) \(\Rightarrow\frac{1}{3}.\frac{3}{5}......\frac{1007}{1009}.\frac{504}{505}\)=\(\frac{504}{505}\)
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Tính A =\(\dfrac{3^2-1}{5^2-1}.\dfrac{7^2-1}{^{ }9^2-1}......\dfrac{2015^2-1}{2017^2-1}.\dfrac{2017^2-1}{2019^2-1}\)
tính
a) left[dfrac{0.8divleft(dfrac{4}{5}cdot1025right)}{0.64-1}+dfrac{left(1.08-dfrac{2}{25}right)divdfrac{4}{7}}{left(6dfrac{5}{7}-3dfrac{1}{4}right)cdot2dfrac{2}{17}}+left(1.2cdot0.5right)divdfrac{4}{5}right]
b) left(0.2right)^{-3}left[left(-dfrac{1}{5}right)^{-2}right]^{-1}+left[left(dfrac{1}{2}right)^{-3}right]^{-2}divleft(2^{-3}right)^{-1}-left(0.175right)^{-2}
c) 2+dfrac{1}{1+dfrac{1}{2+dfrac{1}{1+dfrac{1}{2}}}}
d) dfrac{1}{90}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-dfrac{1}{3}
e) l...
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tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
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d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
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f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)
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Cho A= 1 + \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{4034}\); B = 1 + \(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4033}\)
So sánh \(\dfrac{A}{B}\)với 1\(\dfrac{2017}{2018}\)
Bài 1: Thực hiện phép tính:
Aleft(-dfrac{1}{125}right)^{11}:left(dfrac{1}{5}right)^{32}
B1+dfrac{1}{3}+left(dfrac{1}{3}right)^2+....+left(dfrac{1}{3}right)^{2018}
Cdfrac{16^3cdot3^{10}+120cdot6^9}{4^6cdot3^{12}+6^{11}}
Dleft(dfrac{0.4-dfrac{2}{9}+dfrac{2}{11}}{1.4-dfrac{7}{9}+dfrac{7}{11}}-dfrac{dfrac{1}{3}-0.25+dfrac{1}{5}}{1dfrac{1}{6}-0.875+0.7}right):dfrac{2017}{2018}
Edfrac{1}{2}+dfrac{1}{6}+dfrac{1}{12}+dfrac{1}{20}+dfrac{1}{30}+dfrac{1}{42}+dfrac{1}{56}+dfrac{1}{72}
Gdfrac{left(dfra...
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Bài 1: Thực hiện phép tính:
\(A=\left(-\dfrac{1}{125}\right)^{11}:\left(\dfrac{1}{5}\right)^{32}\)
\(B=1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+....+\left(\dfrac{1}{3}\right)^{2018}\)
\(C=\dfrac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)
\(D=\left(\dfrac{0.4-\dfrac{2}{9}+\dfrac{2}{11}}{1.4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0.25+\dfrac{1}{5}}{1\dfrac{1}{6}-0.875+0.7}\right):\dfrac{2017}{2018}\)
\(E=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(G=\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(2\dfrac{1}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{512+2^7\cdot5^2}:\dfrac{\left(\dfrac{1}{2}\right)^0}{\left(-1\right)^{2017}}\)
Mn ơi giúp e với ........ Em đang cần gấp giúp e với nha!!
Thank you mn nhiều nhiều.....
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
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Tính hoac tinh nhanh neu duoc:
a)\(\left(\dfrac{7}{6}+\dfrac{1}{2}\right):2-2:\left(\dfrac{7}{6}+\dfrac{1}{2}\right)\)
b)\(\left|_{ }-\dfrac{2}{5}+2\right|.\left(\dfrac{1}{3}\right)^3-\dfrac{1}{5}.3+2017^0\)
a: \(=\dfrac{7+3}{6}\cdot\dfrac{1}{2}-2:\dfrac{7+3}{6}\)
\(=\dfrac{10}{12}-2\cdot\dfrac{6}{10}\)
\(=\dfrac{5}{6}-\dfrac{6}{5}=\dfrac{25-36}{30}=-\dfrac{11}{30}\)
b: \(=\left|\dfrac{10}{5}-\dfrac{2}{5}\right|\cdot\dfrac{1}{27}-\dfrac{3}{5}+1\)
\(=\dfrac{8}{5}\cdot\dfrac{1}{27}-\dfrac{3}{5}+1\)
\(=\dfrac{8}{135}-\dfrac{81}{135}+\dfrac{135}{135}=\dfrac{62}{135}\)
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20 tháng 2 2023 lúc 22:10
A=\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+....+\(\dfrac{1}{2017^2}\)
Giải giúp e với ạ. E cảm ơn
Tính giá trị biểu thức sau :
\(B=\left(2017-\dfrac{1}{4}-\dfrac{2}{5}-\dfrac{3}{6}-\dfrac{4}{7}-....-\dfrac{2017}{2020}\right):\left(\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+\dfrac{1}{35}+....+\dfrac{1}{10100}\right)\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)