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\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{99.101}{100^2}\)
\(=\frac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)
\(=\frac{1.2.3.....99}{2.3.4.....100}.\frac{3.4.5.....101}{2.3.4.....100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Ủng hộ mk nha,chúc bn học tốt!!!
\(C=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(C=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{99\cdot101}{100\cdot100}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(C=\frac{1}{100}\cdot\frac{101}{2}\)
\(C=\frac{101}{200}\)
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x......x\frac{9999}{10000}\)
\(C=\frac{1.3}{2^2}x\frac{2.4}{3^2}x\frac{3.5}{4^2}x....x\frac{99.101}{100^2}\)
\(C=\frac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2.......100^2}\)
\(C=\frac{1.2.3.......99}{2.3.4....100}x\frac{3.4.5.....101}{2.3.4......100}\)
\(C=\frac{1}{100}.\frac{101}{2}=\frac{1.101}{100.2}=\frac{101}{200}\)
Ủng hộ mk nha!!!!
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(=\frac{\left(1.2.3....99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(=\frac{1.101}{100.2}=\frac{101}{200}\)
= 3 . 8 . 15 .... 9999 / 4 . 9 . 16 .... 10000
= ( 1 . 3 ) . ( 2 . 4 ) .( 3 . 5) .... ( 99 .... 101 ) / ( 2. 2) . (3.3). (4.4)...(100.100)
= 1. 101/100.2
= 101/ 200
k nha , đúng đó
1*3/2*2.2*4/3*3.3*5/4*4.....99*101/100*100. =1*2*3*...*99/2*3*4*...*100.3*4*5*...*101/2*3*4*...*100. =1/100 . 101/2. =101/200.
3/4 . 8/9 . 15/16 ... 9999/10000
= 1.3/2.2 . 2.4/3.3 ... 99.101/100.100
= 1 . 2 . ... . 99 / 2 . 3 . 100 × 3 . 4 ... 101 / 2 . 3 ... 100
= 1 / 100 . 101 / 2
= 101 / 200
=1.3/2.2 .2.4/3.3 .3.5/4.4 . ...... 99.101/100.100
=1.2.3.4.5 ...... .99/2.3.4.....100 . 3.4.5 ....... .101/2.3.4.5 .... .100
=1/100 .101/2
=101/200
k cho mink nha
3/4.8/9.15/16...9999/10000
=\(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\)...\(\dfrac{99.101}{100.100}\)
=\(\dfrac{1.2...99}{2.3.100}\).\(\dfrac{3.4...101}{2.3.100}\)
=\(\dfrac{1}{100}\).\(\dfrac{101}{2}\)
=\(\dfrac{101}{200}\)
\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}\)+...+\(\frac{9999}{10000}\)
= (1-\(\frac{1}{4}\)) +(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{10000}\))
= 99 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)) => 99 - A
Dễ thấy A>0 =>S < 99 (1)
Lại có A= \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)
=> A<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...\(\frac{1}{99}\)-\(\frac{1}{100}\)
=>A<1-\(\frac{1}{100}\)<1
...
= 3/4 x 8/9 x 15/16 x ... x 9999/10000
= 3 x 8 x 15 x ... x 9999/ 4 x 9 x 16 x ... x 10000
= (1 x 3) x (2 x 4) x (3 x 5) x ... x (99 x 101)/ (2 x 2) x (3 x 3) x (4 x 4) x ... x (100 x 100)
= (1 x 2 x 3 x ... x 99) x (3 x 4 x 5 x ... x 101)/ (2 x 3 x 4 x ... x 100) x (2 x 3 x 4 x ... x 100)
= 1x 101/ 100 x 2
= 101/200
\(\frac{101}{200}\)