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\(B=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(B=-\frac{3}{2^2}.\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right)...\left(-\frac{9999}{100^2}\right)\)
\(B=-\left(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\right)\)(Vì có 99 thừa số, mỗi thừa số là âm nên kết quả là âm)
\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\right)\)
\(B=-\left(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\right)\)
\(B=-\left(\frac{1}{100}.\frac{101}{2}\right)\)
\(B=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
\(\Rightarrow B< -\frac{1}{2}\)
-A =( 1- 1/2 )(1 -1/3).....(1 -1/10)
= 1/2 . 2/3 ..... 9/10
= 1/10
-A = 1/10 nên A = -1/10
Vì 1/10 < 1/9 nên -1/10 > -1/9
Vậy A > -1/9
\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=-\frac{1}{2}.-\frac{2}{3}...-\frac{9}{10}\)
\(=\frac{-\left(1.2...9\right)}{2.3...10}=\frac{-1}{10}\)
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)