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đáp số cuối cùng là 0 vì 1-1/4 = 0/4 = 0 x cho những số nào khác cũng bằng 0 thôi
\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times...\times\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{999}{10000}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times...\times\frac{99\times101}{100\times100}\)
\(=\frac{1}{2}\times\frac{101}{100}\)
\(=\frac{101}{200}\)
\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right).....\left(1-\dfrac{1}{10000}\right)\)
\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)
\(=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4....100.100}\)
\(=\dfrac{\left(1.2.3...99\right)}{2.3.4....100}\cdot\dfrac{3.4.5...101}{2.3.4....100}=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
Ta có: \(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{10000}\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-9999}{10000}\)
\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\)
\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot1111\cdot9}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)
\(=\dfrac{101}{200}\)
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{100}{2}=50\)
`(1-1/4) \times (1-1/9) \times (1-1/16) \times ... \times (1-1/10000)`
`= 3/4 \times 8/9 \times 15/16 \times ... \times 9999/10000`
`= (3 \times 8 \times 15 \times ... \times 9999)/(4 \times 9 \times 16 \times ... \times 10000)`
`=`\(\dfrac{\left(1\times3\right)\times\left(2\times4\right)\times\left(3\times5\right)\times...\times\left(99\times101\right)}{\left(2\times2\right)\times\left(3\times3\right)\times\left(4\times4\right)\times...\times\left(100\times100\right)}\)
`=` \(\dfrac{\left(1\times2\times3\times...\times99\right)\times\left(3\times4\times5\times...\times101\right)}{\left(2\times3\times4\times...\times100\right)\times\left(2\times3\times4\times...\times100\right)}\)
`=` \(\dfrac{1\times101}{2\times100}\)
`= 101/200.`
(1-1/4) x ( 1-1/9) x ( 1- 1/16) x....x (1 - 1/10000)
= 3/4 x 8/9 x 15/16 x ... x 9999/10000
= (1x3/2x2) x ( 2x4/3x3 ) x ( 3 x 5 / 4x4 ) x ... x ( 99 x 101 / 100x 100 )
= \(\dfrac{\left(1.2.3...99\right).\left(3.4.5.101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}\)
= 1 x 101 / 100 x 2
= 101/200