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14 tháng 7 2017

a, Ta có:

\(\frac{1}{2^3}< \frac{1}{1\cdot2\cdot3};\frac{1}{3^3}< \frac{1}{2\cdot3\cdot4};\frac{1}{4^3}< \frac{1}{3\cdot4\cdot5};...;\frac{1}{n^3}< \frac{1}{\left[n-1\right]n\left[n+1\right]}\)

\(\Rightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)

Đặt \(A'=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)

\(\Rightarrow\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{\left[n-1\right].n}-\frac{1}{n\left[n+1\right]}\)

\(\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{n\left[n+1\right]}=\frac{1}{2}-\frac{1}{n\left[n+1\right]}=\frac{1}{4}-\frac{1}{2n\left[n+1\right]}< \frac{1}{4}\)

Vậy \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}< \frac{1}{4}\Leftrightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}< \frac{1}{4}\)

b,

\(C=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}=1+\frac{1}{3}+1+\frac{1}{3^2}+1+\frac{1}{3^3}+...+1+\frac{1}{3^{98}}\)

\(=\left[1+1+1+...+1\right]+\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

Đặt \(C'=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3C'=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\)

\(\Rightarrow3C'-C'=\left[1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right]-\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=1-\frac{1}{3^{98}}\)

\(\Rightarrow C'=\frac{1-\frac{1}{3^{98}}}{2}< 1\)

\(\Rightarrow98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}< 98+1=99< 100\)

\(\Rightarrow\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)

c,

\(D=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{39}}\)

\(4D=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\)

\(4D-D=\left[5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\right]-\left[\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}+\frac{5}{4^{39}}\right]\)

\(3D=5-\frac{5}{4^{39}}\Leftrightarrow D=\frac{5-\frac{5}{4^{39}}}{3}< \frac{5}{3}\)

Vậy:...........

AI THẤY ĐÚNG NHỚ ỦNG HỘ NHA

12 tháng 2 2016

Bạn làm ra đi

10 tháng 5 2016

B = \(\frac{4}{3^1}+\frac{10}{3^2}+\frac{28}{3^3}+...+\frac{3^{98}+1}{3^{98}}\)

B = \(\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{3^3}\right)+...+\left(1-\frac{1}{3^{98}}\right)\)

B = \(\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

B = \(98-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

=> B < 98 < 100

vậy B < 100

10 tháng 5 2016

dễ thui chờ tí nhé

27 tháng 7 2015

OK. Tối nhớ giải hộ mik nha

Mik hứa sẽ lik-e cho bạn

 

26 tháng 2 2017

mình ko biết

11 tháng 5 2016

\(B=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{3^{98}}\right)\)

\(B=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)

\(B=98-\left(\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)<98\)

=>B<98<100

=>B<100 (đpcm)

30 tháng 4 2017

\(1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)\(\frac{1}{3^{98}}\)

\(=1.98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

\(\Rightarrow3A-A=2A=1-\frac{1}{3^{98}}\Rightarrow A=\frac{1-\frac{1}{2^{98}}}{2}< 1\)

\(\Rightarrow B=98+A< 98+1< 99< 100\)

\(\Rightarrow B< 100\)

HQ
Hà Quang Minh
Giáo viên
8 tháng 10 2023

(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai

(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai

(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng

(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} =  - 1\) => D sai

=> Chọn C.

23 tháng 4 2018

Xét \(B=\frac{4}{3}+\frac{10}{9}+...+\frac{3^{98}+1}{3^{98}}\)

   \(\Leftrightarrow B=\frac{3+1}{3}+\frac{9+1}{9}+...+\frac{3^{98}+1}{3^{98}}\)

   \(\Leftrightarrow B=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)(có 98 cặp số hạng)

\(\Leftrightarrow B=\left(1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)(có 98 số hạng 1)

\(\Leftrightarrow B=98+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

Lấy 3A-A, ta được:

\(2A=1-\frac{1}{3^{98}}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2\cdot3^{98}}\)(*)

Thay (*) vào biểu thức B, ta được

\(B=98+\frac{1}{2}-\frac{1}{2\cdot3^{98}}< 100\)

VẬY, B<100 (ĐPCM)

23 tháng 4 2018

Ta có : 

\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)

\(B=\frac{3+1}{3}+\frac{9+1}{9}+\frac{27+1}{27}+...+\frac{3^{98}+1}{3^{98}}\)

\(B=\frac{3}{3}+\frac{1}{3}+\frac{9}{9}+\frac{1}{9}+\frac{27}{27}+\frac{1}{27}+...+\frac{3^{98}}{3^{98}}+\frac{1}{3^{98}}\)

\(B=1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)

\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)

\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

Do từ \(1\) đến \(98\) có \(98-1+1=98\) số hạng nên có \(98\) số \(1\) suy ra : 

\(B=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\) ta có : 

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

\(2A=1-\frac{1}{3^{98}}< 1\)

Mà \(2A< 1\)\(\Rightarrow\)\(A< 1\)

Do đó : 

\(B=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)< 98+1=99< 100\)

\(\Rightarrow\)\(B< 100\) ( đpcm ) 

Vậy \(B< 100\)

Chúc bạn học tốt ~