= 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
= 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
Chứng minh rằng: \(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}>98\)
Cho C = \(\dfrac{3}{4}\) +\(\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\)
Chứng minh rằng C>98
Bài 1:Tính
a, A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot....\cdot\dfrac{9999}{10000}\)
b,B=\(\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot....\cdot\left(1-\dfrac{1}{1326}\right)\)
c,C=\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)
CMR C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\)
Bài 1: Thực hiện phép tính:
a, \(\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right)\)
b, \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\)
c, \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\)
bài 4 : cmr :
c) C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)
\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}._{......}.\dfrac{80}{81}.\dfrac{99}{100}\)
tính nhanh
A=\(\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}...\dfrac{899}{30^2}\)
B=\(\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}...\dfrac{2499}{2500}\)
bài 5
cho A=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\dfrac{5}{6}\cdot...\cdot\dfrac{9999}{10000}\)
So sánh a với \(\dfrac{1}{100}\)