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26 tháng 4 2021

Thanks trước

helppp me?

BT1: CMR: a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\) c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\) d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\) e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\) f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\) BT2: Tính tổng a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\) b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\) BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\) CMR: 1 < S <...
Đọc tiếp

BT1: CMR:

a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)

b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)

c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)

d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)

f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)

BT2: Tính tổng

a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)

BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)

CMR: 1 < S < 2

1
22 tháng 3 2017

bài này có trong sách Nâng cao và Phát triển bạn nhé

Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)

\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)

\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)

a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)

b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)

\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)

c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)

\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)

d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)

\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)

a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)

\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)

\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)

b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)

\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)

\(=\dfrac{-5}{3}+\dfrac{42}{29}\)

\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)

c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)

\(=-1-1-1+4\)

=1

18 tháng 7 2021

a) Ta có: −518+3245−910−518+3245−910

=−2590+6490−8190=−2590+6490−8190

=−4290=−715=−4290=−715

b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)

=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229

=−53+4229=−53+4229

=−14587+12687=−1987=−14587+12687=−1987

c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1

=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4

=−1−1−1+4=−1−1−1+4

=1

bạn ơi cái câu <1 số hạng cuối cùng là j thế?

6 tháng 5 2021

chỉ thế thôi nha bạn

 

18 tháng 9 2022

Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
                                         = 1/2-1/3+1/3-1/4+...+1/10-1/11 

                                         = 1/2 - 1/11 = 9/22  (đpcm)