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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\(\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{4}\right)^5\cdot3}{\dfrac{1}{1024}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot2}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)
\(=2:\dfrac{-1}{6}=2\cdot\left(-6\right)=-12\)
\(T=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{10}\right)\)\(\Rightarrow T=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.\dfrac{8}{9}.\dfrac{10}{11}.\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
\(\Rightarrow=\dfrac{1}{11}\)
\(\Rightarrow\) Số nghịch đảo của T là \(11\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
a: =>4y+15/16=1
=>4y=1/16
hay y=1/64
b: =>10y+1023/1024=1
=>10y=1/1024
hay y=1/10240
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ =\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ =\dfrac{2}{-\dfrac{1}{6}}\\ =-12\)
\(B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot5-\left(\dfrac{1}{2}\right)^{10}\cdot3}{\left(\dfrac{1}{2}\right)^{10}\cdot\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{11}}\\ B=\dfrac{\left(\dfrac{1}{2}\right)^{10}\cdot\left(5-3\right)}{\left(\dfrac{1}{2}\right)^{10}\cdot\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\\ B=\dfrac{2}{-\dfrac{1}{6}}\\ B=-12\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
nhầm đề
phá ngoặc kiểu j vậy bn