So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
Cho A = 1 - \(\frac{1}{2}\) +\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+...+\(\frac{1}{149}\)-\(\frac{1}{150}\) ; B = \(\frac{1}{76}\)+\(\frac{1}{77}\)+\(\frac{1}{78}\)+...+\(\frac{1}{150}\).
Tính \(\frac{A}{B}\)
A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/149 - 1/150
A = (1 + 1/3 + 1/5 + ... + 1/149) - (1/2 + 1/4 + 1/6 + ... + 1/150)
A = (1 + 1/2 + 1/3 +1/4 + 1/5 + 1/6 + ... + 1/149 + 1/150 - 2.(1/2 + 1/4 + 1/6 + ... + 1/150)
A = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/149 + 1/150) - (1 + 1/2 + 1/3 + ... + 1/75)
A =1/76 + 1/77 + 1/78 + ... + 1/150
=> A/B = 1
Cho A = 1 - \(\frac{1}{2}\) +\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+...+\(\frac{1}{149}\)-\(\frac{1}{150}\) ; B = \(\frac{1}{76}\)+\(\frac{1}{77}\)+\(\frac{1}{78}\)+...+\(\frac{1}{150}\).
Tính \(\frac{A}{B}\)
A = \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}-1\right)\times\left(1-\frac{8}{1}-\frac{4}{1}-\frac{2}{1}\right)\)
B = \(\frac{\frac{3}{1}-\frac{6}{3}-\frac{9}{6}-\frac{369}{1}}{\frac{1}{3}+\frac{3}{6}+\frac{6}{9}-\frac{1}{963}}\)
C = \(\frac{1}{1}-\frac{1}{2}+\frac{3}{1}-\frac{1}{4}+\frac{5}{1}-\frac{1}{6}+\frac{7}{1}-\frac{1}{8}+\frac{9}{1}-\frac{1}{10}\)
so sánh các số trên ( A , B , C )
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
**************************l i k e***********************************8
A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
Bài 1:CMR:\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}<2\)
Bài 2: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right)}=\frac{99}{101}\)
Bài 3:\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2449}{2500}\)
Bài 4:CMR:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng tỏ rằng:
\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{149}+\frac{1}{150}>\frac{5}{6}\)
Chứng minh rằng :
a, \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}<\frac{1}{12}\)
b, \(\frac{1}{6}<\frac{1}{^{5^2}}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{4}\)
c, \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
d, \(1<\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}<2\)
b.Đặt A = \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+....+\frac{1}{100^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)= \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\)= \(\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}\frac{1}{6}\)(2)
Từ (1) và (2) =>\(\frac{1}{6}\) < A < \(\frac{1}{4}\)
So sánh \(A\)với\(13\),biết rằng:
\(A=\frac{13}{15}+\frac{7}{5}+\frac{3}{4}+\frac{1}{5}+\frac{1}{7}+\frac{19}{20}+\frac{5}{4}+\frac{1}{3}+\frac{1}{6}+\frac{1}{13}+\frac{17}{23}+\frac{9}{8}+\frac{2}{5}+\frac{1}{7}+\frac{1}{25}+\frac{3}{2}+\frac{1}{8}+\frac{1}{19}+\frac{1}{9}+\frac{1}{97}\)
1/ Chứng tỏ rằng \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}<1\)
2/ Chứng tỏ rằng \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}<1\)
3/ Rút gọn biểu thức \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
4/ Tính nhanh\(\frac{\frac{4}{2010}+\frac{4}{2011}-\frac{4}{2012}}{\frac{5}{2010}+\frac{5}{2011}-\frac{5}{2012}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
GIÚP ĐƯỢC CÂU NÀO THÌ GIÚP NHÉ, MÌNH TICK CHO
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
a)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)=\(\frac{99}{100}<1\)
