chợ \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{1997.1998}\)
Va \(B=\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998.1000}\)
CHUNG MINH RANG A/B LA SO NGUYEN.
Những câu hỏi liên quan
Cho: A=1/1.2 + 1/3.4 + 1/5.6 +...+ 1/1997.1998; B= 1/1000.1998+1/1001.1997+1/1002.1996 +...+ 1/1998.1000
Chứng minh rằng:A:B là một số nguyên.
A = 1/(1.2) + 1/(3.4) + 1/(5.6) +....+ 1/(1997.1998) =
(1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 1997 - 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - 2(1 / 2 + 1 / 4 + ... + 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - (1 + 1 / 2 + ... + 1 / 999) =
1 / 1000 + 1 / 1001 + ... + 1 / 1998
2A = (1 / 1000 + 1 / 1001 + ... + 1 / 1998) + (1 / 1998 + 1 / 1997 + ... + 1 / 1000) =
(1 / 1000 + 1 / 1998) + (1 / 1001 + 1 / 1997) + ... + (1 / 1998 + 1 / 1000) =
2998*[1 / (1000*1998) + 1 / (1001*1997) + ... + 1 / (1998*1000)] = 2998B
=> A / B = 1499 nguyên
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A = (1/1.2) + (1/3.4) + (1/5.6) +....+ ( 1/1997.1998)
ta có
1/1*2 = 1 - 1/2
1/3*4 = 1/3 - 1/4
...
1/1997*1998 = 1/1007 - 1/1998
bạn gộp lại tự giải tiếp nha
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Áp dụng 1 / [n(n+1)] = 1 / n - 1 / (n+1) với mọi n ≥ 1 có:
A = 1/(1.2) + 1/(3.4) + 1/(5.6) +....+ 1/(1997.1998) =
(1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 1997 - 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - 2(1 / 2 + 1 / 4 + ... + 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - (1 + 1 / 2 + ... + 1 / 999) =
1 / 1000 + 1 / 1001 + ... + 1 / 1998
2A = (1 / 1000 + 1 / 1001 + ... + 1 / 1998) + (1 / 1998 + 1 / 1997 + ... + 1 / 1000) =
(1 / 1000 + 1 / 1998) + (1 / 1001 + 1 / 1997) + ... + (1 / 1998 + 1 / 1000) =
2998*[1 / (1000*1998) + 1 / (1001*1997) + ... + 1 / (1998*1000)] = 2998B
=> A / B = 1499 nguyên
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\(\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998.1000}\)
\(\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998+1000}\)
\(S=\frac{1}{1000.1998}+\frac{1}{1001.1997}+...+\frac{1}{1998.1000}\)
\(=\frac{1}{2998}\left(\frac{1000+1998}{1000.1998}+\frac{1001+1997}{1001.1997}+...+\frac{1998+1000}{1998.1000}\right)\)
\(=\frac{1}{2998}\left(\frac{1}{1000}+\frac{1}{1998}+\frac{1}{1001}+\frac{1}{1997}+...+\frac{1}{1998}+\frac{1}{1000}\right)\)
\(=\frac{2}{2998}\left(\frac{1}{1000}+\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{1998}\right)\)
\(=\frac{1}{1499}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1998}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}\right)\right]\)
\(=\frac{1}{1499}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1998}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1998}\right)\right]\)
\(=\frac{1}{1499}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1997}-\frac{1}{1998}\right)\)
\(=\frac{1}{1499}\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{1997.1998}\right)\)
Afrac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+....+frac{1}{99.100}&Bfrac{2014}{51}+frac{2018}{52}+frac{2018}{53}+....+frac{2018}{100}Chung minh minh rang frac{B}{A}la 1 so nguyen.Ai làm đc mình dùng 2nick,mình tick cho
Đọc tiếp
A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+....+\(\frac{1}{99.100}\)&B=\(\frac{2014}{51}\)+\(\frac{2018}{52}\)+\(\frac{2018}{53}\)+....+\(\frac{2018}{100}\)
Chung minh minh rang \(\frac{B}{A}\)la 1 so nguyen.
Ai làm đc mình dùng 2nick,mình tick cho
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2.\frac{1}{2}-2.\frac{1}{4}-2.\frac{1}{6}-...-2.\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B=\frac{2018}{51}+\frac{2018}{52}+\frac{2018}{53}+...+\frac{2018}{100}\)
\(=2018.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2018\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}\)
\(=2018\)
Vậy \(\frac{B}{A}\)là 1 số nguyên
!!!
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cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
chung minh rang
a) A =\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
b) \(\frac{25}{75}+\frac{25}{100}< A< \frac{25}{51}+\frac{25}{75}\)
cho a, b, c, d la 4 so nguyen duong thoa man: b= \(\frac{a+c}{2}va\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{b}+\frac{1}{d}\right)\)
chung minh: \(\frac{a}{b}=\frac{c}{d}\)
Chung minh rang:
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
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bn ơi bn có thê
rhuowngs dẫn mình
làm ko vì
mai mình ucngx
có bài này
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1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
nhé !
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Xem thêm câu trả lời
Cho a,b.c la cac so duong va abc = 1
Chung minh rang \(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\)
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
Cần gấp, ai nhanh mik tick nha
Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
Cho 4 so nguyen duong a,b,c,d ma trong do b la trung binh cong cua 2 so a va c.Chung minh rang 4 so a,b,c,d lap thanh mot ti le thuc neu \(\frac{2}{c}\)=\(\frac{1}{b}+\frac{1}{d}\)
Ta có : \(b=\frac{a+c}{2}\) \(\implies\) \(2b=a+c\)
\(\frac{2}{c}=\frac{1}{b}+\frac{1}{d}\)
\(\implies\) \(\frac{1}{2}.\frac{2}{c}=\frac{1}{2}.\left(\frac{1}{b}+\frac{1}{d}\right)\)
\(\implies\) \(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{b}+\frac{1}{d}\right)\)
\(\iff\) \(\frac{1}{c}=\frac{b+d}{2db}\)
\(2db=c.\left(b+d\right)\)
\(\left(a+c\right)d=cd+cb\)
\(ad+cd=cd+cb\)
\(ad=cb\)
\(\frac{a}{c}=\frac{b}{d}\) là một tỉ lệ thức \(\left(đpcm\right)\)