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PT
0
22 tháng 10 2016
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{121}-1\right)\)
= \(-\frac{3}{4}.-\frac{8}{9}...-\frac{120}{121}\)
= \(\frac{-3.8....120}{4.9...121}\)
= \(\frac{-1.3.2.4...10.12}{2^2.3^3...11^2}\)
= \(-\frac{12}{2.11}\)
= \(-\frac{6}{11}\)
22 tháng 10 2016
Tính
(1/4-1).(1/9-1).(1/16-1).....(1/121-1)
\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).....\left(\frac{1}{121}-1\right)\\ =\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-120}{121}\)
\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-10.12}{11.11}\)
\(=\frac{\left(-1.3\right).\left(-2.4\right).\left(-3.5\right).....\left(-10.12\right)}{2^2.3^2.4^2.....11^2}\\ =\frac{\left(-1\right).\left(-2\right).\left(-3\right).....\left(-10\right)}{2.3.4.....11}.\frac{3.4.5.....12}{2.3.4.....11}\)
\(=\frac{1}{2}.\frac{12}{11}\\ =\frac{6}{11}\)
HA
22 tháng 3 2017
a,
\(\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}........\dfrac{-99}{100}.\dfrac{-120}{121}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.........\dfrac{9.11}{10^2}.\dfrac{10.12}{11^2}\)
\(=\dfrac{1.2.3.4.....10.3.4.5.6......11.12}{2^2.3^2........11^2}\)
\(=\dfrac{1.2.11.12}{2^2.11^2}=\dfrac{12}{22}\)
YN
22 tháng 3 2017
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\\ \Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(M=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow S=2^{2010}-M\)
* Tính M
\(M=2^{2009}+2^{2008}+...+2+1\\ \Rightarrow2^0+2^1+...+2^{2008}+2^{2009}\\ \Rightarrow2S=2^1+2^2+...+2^{2009}+2^{2010}\\ \Rightarrow2S-S=\left(2^1+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\\ \Rightarrow S=2^{2010}-2^0=2^{2010}-1\)Thay M vào S, ta được :
\(S=2^{2010}-\left(2^{2010}-1\right)\\ \Rightarrow S=2^{2010}-2^{2010}+1\\ \Rightarrow S=1\)
\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....\left(\frac{1}{101}-1\right)\)
\(=\frac{3}{4}.\frac{8}{9}.....\frac{120}{121}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{10.12}{11.11}\)
\(=\frac{\left(1.3\right).\left(2.4\right).....\left(10.12\right)}{\left(2.2\right).\left(3.3\right).....\left(11.11\right)}\)
\(=\frac{\left(1.2.3.....10\right).\left(3.4.5.....12\right)}{\left(2.3.4.....11\right).\left(2.3.4.....11\right)}\)
\(=\frac{1.12}{11.2}=\frac{6}{11}\)