T=\(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{512}\right)+\left(1-\dfrac{1}{1024}\right)\)
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Tính nhanh
A=\(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{1024}\right)\)
B=4.5100 .\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\right)+1\)
\(\dfrac{\left(\dfrac{1}{2}\right)^{10}.5-\left(\dfrac{1}{4}\right)^5.3}{\dfrac{1}{1024}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
\(\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{4}\right)^5\cdot3}{\dfrac{1}{1024}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot2}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)
\(=2:\dfrac{-1}{6}=2\cdot\left(-6\right)=-12\)
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cho G = \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)+\left(1+\dfrac{1}{256}\right)......\left(1+\dfrac{1}{2^{1024}}\right)\)và H = \(\dfrac{1}{2^{2047}}\)
Tính G + H
Tìm tích sau rồi tìm số nghịch đảo của kết quả :
\(T=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{10}\right)\)
\(T=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{10}\right)\)\(\Rightarrow T=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.\dfrac{8}{9}.\dfrac{10}{11}.\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
\(\Rightarrow=\dfrac{1}{11}\)
\(\Rightarrow\) Số nghịch đảo của T là \(11\)
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\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
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21 tháng 3 2022 lúc 8:43
Bài 5. Tìm \(y\) biết:
a) \(\left(y+\dfrac{1}{2}\right)+\left(y+\dfrac{1}{4}\right)+\left(y+\dfrac{1}{8}\right)+\left(y+\dfrac{1}{16}\right)=1\)
b) \(\left(y+\dfrac{1}{2}\right)+\left(y+\dfrac{1}{4}\right)+\left(y+\dfrac{1}{8}\right)+...+\left(y+\dfrac{1}{1024}\right)=1\)
a: =>4y+15/16=1
=>4y=1/16
hay y=1/64
b: =>10y+1023/1024=1
=>10y=1/1024
hay y=1/10240
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tính giá trị của biểu thức : B\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}.5-\left(\dfrac{1}{4}\right)^5.3}{\dfrac{1}{1024}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)
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\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)
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Tính các tích sau:
P_1 left(1+dfrac{2}{4}right)left(1+dfrac{2}{10}right)left(1+dfrac{2}{18}right)....left(1+dfrac{2}{n^2+3n}right)
P_2 left(1+dfrac{1}{3}right)left(1+dfrac{1}{8}right)left(1+dfrac{1}{15}right)....left(1+dfrac{2}{n^2+2n}right)
P_3 left(1-dfrac{1}{1+2}right)left(1-dfrac{1}{1+2+3}right)left(1-dfrac{1}{1+2+3+4}right).....left(1-dfrac{1}{1+2+3+4+...+n}right)
P_4 dfrac{2^4+4}{4^4+4}.dfrac{6^4+4}{8^4+4}.dfrac{8^4+4}{10^4+4}....dfrac{18^4+4}{20^4+4}
Đọc tiếp
Tính các tích sau:
P\(_1\) =\(\left(1+\dfrac{2}{4}\right)\left(1+\dfrac{2}{10}\right)\left(1+\dfrac{2}{18}\right)....\left(1+\dfrac{2}{n^2+3n}\right)\)
P\(_2\) =\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{2}{n^2+2n}\right)\)
P\(_3\) = \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right).....\left(1-\dfrac{1}{1+2+3+4+...+n}\right)\)
P\(_4\) = \(\dfrac{2^4+4}{4^4+4}.\dfrac{6^4+4}{8^4+4}.\dfrac{8^4+4}{10^4+4}....\dfrac{18^4+4}{20^4+4}\)
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
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