Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\left(9\text{số (-1)}\right)\right].\frac{3}{4}.\frac{8}{9}....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right)\right].\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
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
$A=\left(\dfrac14-1\right)\left(\dfrac19-1\right)\left(\dfrac1{16}-1\right)\cdots\left(\dfrac1{100}-1\right)\left(\dfrac1{121}-1\right)$
$=\left(-\dfrac34\right)\left(-\dfrac89\right)\left(-\dfrac{15}{16}\right)\cdots\left(-\dfrac{99}{100}\right)\left(-\dfrac{120}{121}\right)$
$=(-1)^{10}\cdot\dfrac34\cdot\dfrac89\cdot\dfrac{15}{16}\cdots\dfrac{99}{100}\cdot\dfrac{120}{121}$
$=\dfrac34\cdot\dfrac89\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdots\dfrac{99}{100}\cdot\dfrac{120}{121}$
$=\dfrac{3\cdot8\cdot15\cdot24\cdots99\cdot120}{4\cdot9\cdot16\cdot25\cdots100\cdot121}$
$=\dfrac{(1\cdot2\cdot3\cdots10)\,(3\cdot4\cdot5\cdots12)}{(2\cdot3\cdot4\cdots11)^2}$
$=\dfrac{1\cdot12}{2\cdot11}$
$=\dfrac6{11}.$
\(A=\left(\frac{1-2^2}{2^2}\right)\left(\frac{1-3^2}{3^2}\right)....\left(\frac{1-10^2}{10^2}\right)\)
\(A=\frac{\left(1+2\right)\left(1-2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.......\frac{\left(1-10\right)\left(1+10\right)}{10^2}\)
\(A=\frac{3.\left(-1\right)}{2^2}.\frac{\left(-2\right).4}{3^2}........\frac{\left(-9\right).11}{10^2}=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{9.11}{10^2}\right)\)
\(=-\left(\frac{1.2....9}{2.3....10}.\frac{3.4....11}{2.3.4...10}\right)=-\left(\frac{1}{10}.\frac{11}{2}\right)=\frac{-11}{20}\)