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A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)
A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)
Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)
nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)
\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)
A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)
A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)
\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\)
Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A> \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)
Tương tự ta có :
A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)
A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)
A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)
\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)
cho a =1/2.3/4.5/6.....99/100.Chứng minh rằng:1/15<a<1/10.
ta co a < 2/3.4/5.....100/101
nhan hai ve cho a ta co
a^2 <2/3.4/5...100/101.1/2.3/4.5/6...99/100
a^2<1/101 <1/100
a< can 1/100 a <1/10.
Cm tương tự ta dc a>1/15.
Bn cx có thể kham khảo bài làm khác là:https://diendan.hocmai.vn/threads/toan-6-cmr-a-1-10-va-a-1-15.223994/
Đặt A = \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{2001}{3^{2001}}\)
3A = \(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{2001}{3^{2000}}\)
3A - A = ( \(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{2001}{3^{2000}}\) ) - ( \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{2001}{3^{2001}}\) )
2A = 1 + \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2000}}-\dfrac{2001}{3^{2001}}\)
Đặt B = 1 + \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2000}}\)
3B = 3 + 1 + \(\dfrac{1}{3}+...+\dfrac{1}{3^{1999}}\)
3B - B = ( 3 + 1 + \(\dfrac{1}{3}+...+\dfrac{1}{3^{1999}}\) ) - ( 1 + \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2000}}\) )
2B = 3 - \(\dfrac{1}{3^{2000}}\) -
B = \(\dfrac{3}{2}-\dfrac{1}{3^{2020}\cdot2}\)
Vậy 2A = \(\dfrac{3}{2}-\dfrac{1}{3^{2000}\cdot2}\) - \(\dfrac{2001}{3^{2001}}\)
A = \(\dfrac{3}{4}-\dfrac{1}{3^{2000}\cdot2^2}-\dfrac{1}{3^{2001}\cdot2}< \dfrac{3}{4}\)
Mà \(\dfrac{3}{4}< \dfrac{4}{5}\)
Vậy A \(< \dfrac{4}{5}\)