Ôn tập toán 6
Các câu hỏi tương tự
cm 3/4+8/9+15/16+.....+9999/10000<99
Chứng minh rằng: \(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}>98\)
Cho \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\).Kết quả rút gọn của 200.A là
Cho C = \(\dfrac{3}{4}\) +\(\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\)
Chứng minh rằng C>98
Tính:3/4 x 8/9 x 15/16 x ..............9999/10000 Bài giải...........................................................................................................................................................................................................................................................................................................................................................................................................................................
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Tính:
3/4 x 8/9 x 15/16 x ..............9999/10000
Bài giải
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\(P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{9999}{10000}\)
đề cương toán có câu hỏi khó của trường thcs hiến thành,mong giải giùm
Bài 1:Tính
a, Adfrac{3}{4}cdotdfrac{8}{9}cdotdfrac{15}{16}cdot....cdotdfrac{9999}{10000}
b,Bleft(1-dfrac{1}{21}right)cdotleft(1-dfrac{1}{28}right)cdotleft(1-dfrac{1}{36}right)cdot....cdotleft(1-dfrac{1}{1326}right)
c,Cleft(1+dfrac{1}{1cdot3}right)cdotleft(1+dfrac{1}{2cdot4}right)cdotleft(1+dfrac{1}{3cdot5}right)cdot....cdotleft(1+dfrac{1}{99cdot101}right)
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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=\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)
\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}._{......}.\dfrac{80}{81}.\dfrac{99}{100}\)
Tính nhanh:
a, M=3/1*2+3/2*3+3/3*4+........+3/20*21
b, N=4/3*9/8*16/15*25/24*.......*100/99