M=1/4(4/1*5+8/5*13+...+16/25*41)
=1/4(1-1/5+1/5-1/13+...+1/25-1/41)
=40/41*1/4=10/41
\(N=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{16}+...+\dfrac{1}{43}-\dfrac{1}{61}\right)=\dfrac{1}{3}\cdot\dfrac{60}{61}=\dfrac{20}{61}\)
=>M<N
cho M =1/1.25 + 2/5.13 + 3/13.25 + 4/25.41 và N = 2/1.7 + 3/7.16 + 4/16.28 + 5/28.43. So sánh M và N
So sánh các phân số sau
\(a,\dfrac{-7}{6}và\dfrac{-11}{9}\) b,\(\dfrac{5}{-7}và\dfrac{-4}{5}\)
c,\(\dfrac{-8}{7}và\dfrac{-2}{5}\) d,\(\dfrac{-2}{5}và\dfrac{1}{3}\)
So sánh C và \(\dfrac{1}{100}\) biết: C= \(\dfrac{1}{2}\). \(\dfrac{3}{4}\). \(\dfrac{5}{6}\) . .... . \(\dfrac{9999}{10000}\)
Câu 5 : A= \(\dfrac{1}{2}\) +\(\dfrac{1}{2^2}\)+ \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)+ ....+\(\dfrac{1}{2^{2021}}\)+\(\dfrac{1}{2^{2022}}\)và B= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{17}{60}\)
a) Rút gọn A
b) So sánh A và B
1. Cho N=\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\)
CMR \(\dfrac{3}{5}< N< \dfrac{4}{5}\)
2. Cho M=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{29}{3^{29}}-\dfrac{30}{3^{30}}\)
CMR \(M< \dfrac{3}{16}\)
3. Cho Q=\(\dfrac{2}{3}+\dfrac{8}{9}+\dfrac{26}{27}+...+\dfrac{3^{2021}-1}{3^{2021}}\)
CMR \(Q>\dfrac{4041}{2}\)
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
\(l,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)
\(m,\left(3-2\dfrac{1}{3}+\dfrac{1}{4}\right):\left(4-5\dfrac{1}{6}+2\dfrac{1}{4}\right)\)
\(n,F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(p,F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
Cho A = \(\dfrac{1}{2}x\dfrac{3}{4}x\dfrac{5}{6}x...x\dfrac{99}{100};B=\dfrac{1}{10}\) So sánh: A và B.
