Cho A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 +.....+ 2/97.99 và B = 1^2 / 1.2 x 2^2/2.3 x 3^2 / 3.4 x 4^2 /4.5 x .... x 98^2 / 98.99. Chứng tỏ A = 98B
Những câu hỏi liên quan
A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/98.99+1/99.100
B=2/1.3+2/3.5+2/5.7+2/7.9+...+2/97.99+2/99.101
C=1/2+1/4+1/8+1/16+...+1/1024+1/2048
D=1/2+1/6+1/18+1/54+1/4374+1/13122 ( . (DẤU CHẤM LÀ ''NHÂN'')
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)
\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)
Đúng 0
Bình luận (0)
Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
Đúng 0
Bình luận (0)
Bài: Cho A frac{2}{1.3}+frac{2}{3.5}+frac{2}{5.7}+frac{2}{7.9}+ ... + frac{2}{97.99} và B frac{1^2}{1.2}.frac{2^2}{2.3}.frac{3^2}{3.4}.frac{4^2}{4.5}.....frac{98^2}{98.99}Chứng tỏ: A 98BLưu ý: Đề toán nâng cao lớp 6 HK II của quận 4. Mong mọi người giúp đỡ!
Đọc tiếp
Bài:
Cho A = \(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+ ... + \(\frac{2}{97.99}\) và B =\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\).....\(\frac{98^2}{98.99}\)
Chứng tỏ: A = 98B
Lưu ý: Đề toán nâng cao lớp 6 HK II của quận 4. Mong mọi người giúp đỡ!
+)Ta có:\(A=\frac{2}{1.3}+\frac{2}{3.5}+............+\frac{2}{97.99}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{99}=\frac{99}{99}-\frac{1}{99}=\frac{98}{99}\)
+)Ta lại có:\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...............\frac{98^2}{98.99}\)
\(\Rightarrow B=\frac{\left(1.2.3......98\right).\left(1.2.3.........98\right)}{\left(1.2.3........98\right).\left(2.3.4..........99\right)}\)
\(\Rightarrow B=\frac{1}{99}\left(2\right)\)
+)Từ (2)
\(\Rightarrow98B=98.\frac{1}{99}=\frac{98}{99}=A\)
Hay 98B=A
Vậy 98B=A
Chúc bn học tốt
Xem thêm câu trả lời
tính giá trị của biểu thức
a) A=\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + ...+\(\frac{1}{99.100}\)
b) B= \(\frac{2}{1.3}\)+\(\frac{2}{3.5}\) + \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\) +...+\(\frac{2}{97.99}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
Đúng 0
Bình luận (1)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
Đúng 0
Bình luận (3)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(\frac{1}{1}-\frac{1}{100}\)
=>\(\frac{99}{100}\)
B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
=>\(\frac{1}{1}-\frac{1}{99}\)
=>\(\frac{98}{99}\)
Đúng 0
Bình luận (0)
text{Tìm x, biết:}a) x-dfrac{2}{3.5}-dfrac{2}{5.7}-dfrac{2}{7.9}-dfrac{2}{9.11}-dfrac{2}{11.13}-dfrac{2}{13.15}dfrac{2}{5}b) dfrac{1}{2.3}.x+dfrac{1}{3.4}.x+dfrac{1}{4.5}.x+...+dfrac{1}{49.50}.x1c) x-dfrac{20}{11.3}-dfrac{20}{13.15}-...-dfrac{53}{55}dfrac{3}{11}d) x+dfrac{4}{5.9}+dfrac{4}{9.13}+...+dfrac{4}{41.45}dfrac{-37}{45}e) left(dfrac{11}{12}.dfrac{11}{2.23}.dfrac{11}{23.34}...dfrac{11}{89.100}right).xdfrac{5}{3}f) left(dfrac{2}{11.13}.dfrac{2}{13.15}.dfrac{2}{15.17}...dfrac{2}{19.21}right...
Đọc tiếp
\(\text{Tìm x, biết:}\)
\(a\)) \(x-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-\dfrac{2}{9.11}-\dfrac{2}{11.13}-\dfrac{2}{13.15}=\dfrac{2}{5}\)
\(b\)) \(\dfrac{1}{2.3}.x+\dfrac{1}{3.4}.x+\dfrac{1}{4.5}.x+...+\dfrac{1}{49.50}.x=1\)
\(c\)) \(x-\dfrac{20}{11.3}-\dfrac{20}{13.15}-...-\dfrac{53}{55}=\dfrac{3}{11}\)
\(d\)) \(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\)
\(e\)) \(\left(\dfrac{11}{12}.\dfrac{11}{2.23}.\dfrac{11}{23.34}...\dfrac{11}{89.100}\right).x=\dfrac{5}{3}\)
\(f\)) \(\left(\dfrac{2}{11.13}.\dfrac{2}{13.15}.\dfrac{2}{15.17}...\dfrac{2}{19.21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
Đúng 1
Bình luận (0)
1) Tính 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13+2/13.15+2/1.2+2/2.3+2/3.4+2/4.5+...+2/9.10
2) Tìm x biết: (11/12+11/12.24+11/23.34+...+11/89.100)
1)
2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10
=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)
= (2/3-2/15)+ [2(1-1/10)]
=8/15+9/5
=7/3
2)
11/12+11/12.24+...+11/88.99
=11-1/9
=10/8/9
Đúng 0
Bình luận (0)
23 tháng 4 2021 lúc 13:31
a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)
tinh tong A và B biết:A=1.2+2.3+3.4+4.5+5.6+...+98.99;B=1^2+2^2-3^2+4^2+....+98^2
Bài 5 :
a) Chứng minh rằng : 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/199.200/ 1/101 + 1/102 + 1/103 + ... + 1/200 = 1
b) So sánh A = 1 mũ 2/1.2 x 2 mũ 2/2.3 x 3 mũ 2/3.4 x 99 mũ 2/99.100 x 100 mũ 2/100.101 và B = 2 mũ 2/1.3 x 3 mũ 2/2.4 x 4 mũ 2/3.5
x .... x 59 mũ 2/58.60
Chỗ 4 mũ 2/3.5 x ... x 59 mũ 2/58.60 nha
Đúng 0
Bình luận (0)
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
Đúng 0
Bình luận (0)
A=1.2+2.3+3.4+4.5+...+98.99
B=1^2+2^2+3^2+4^2+...+98^2
tính A-B
Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +.....+ 98.99
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... +98.99.(100 - 97)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100
=> 3A = 98.99.100
=> A = 98.99.100 / 3
=> A = 323400
Đúng 0
Bình luận (0)