cho \(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)\)
so sánh A với \(-\frac{1}{2}\)
Những câu hỏi liên quan
CHO A=\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)\). HÃY SO SÁNH A VỚI -1/2
\(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)
So sánh A với \(-\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)
=> \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{100^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{100^2-1}{100^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=\frac{1.2....99}{2.3....100}.\frac{3.4....101}{2.3....100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
=> \(A=-\frac{101}{200}< -\frac{1}{2}\)
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Cho A = \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
So sánh A với \(-\frac{1}{2}\)
Cho A = \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\cdot\left(\frac{1}{2018^2}-1\right)\) và B = \(-\frac{1}{2}\)
Hãy so sánh A và B
tính nhanh a, frac{-2}{5}cdotleft(frac{5}{17}-frac{9}{15}right)-frac{2}{5}cdotfrac{2}{17}+frac{-2}{5}b, frac{1}{5}cdotleft(frac{4}{13}-frac{9}{11}right)+frac{1}{3}left(frac{9}{13}-frac{4}{22}right)c, left(frac{1}{2}+1right)cdotleft(frac{1}{3}+1right)cdotleft(frac{1}{4}+1right)cdot...cdotleft(frac{1}{99}+1right)d, left(1-frac{1}{2}right)cdotleft(1-frac{1}{3}right)cdotleft(1-frac{1}{4}right)cdot...cdotleft(1-frac{1}{100}right)
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tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Mk ko biết lm nhưng cứ k thoải mái nha
SORRY
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Tính các tích sau: với n là số tự nhiên, n<3
a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{n}\right)\)
b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
\(A=\left(1-\frac{1}{2^2}^{ }^{ }\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{100^2}\right).\)
ai nhanh mình tick
\(B=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{1+2+3+....+100}\right)\)
Tính A=\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)\)trả lời nhanh mình tk nha