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8 tháng 3 2017

Ta co

+)A=2/60*63+2/63*66+...+2/117*120+2/2003

A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003

A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003

A*3/2=1/60-1/120+3/2003

A=(1/120+3/2003)*2/3

+)B=5/40*44+5/44*48+...+5/76*80+5/2003

B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003

B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003

B*4/5=1/40-1/80+4/2003

B=(1/80+4/2003)*5/4

Tu tren ta co A=(1/120+3/2003)*2/3

B=(1/80+4/2003)*5/4

Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)

8 tháng 3 2017

Đáp án này chắc chắn đúng k bạn ơi?

b: \(A=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2003}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{2}{2003}\)

\(=2\left(\dfrac{1}{360}+\dfrac{1}{2003}\right)\)

\(B=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2003}\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2003}\)

\(=5\left(\dfrac{1}{320}+\dfrac{1}{2003}\right)\)

Vì 1/360+1/2003<1/320+1/2003

nên A<B

17 tháng 2 2017

Ta có: \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)

\(\Rightarrow A=\frac{2}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2003}\)

\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2003}\)

\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2003}\)

\(\Rightarrow A=\frac{2}{3}.\frac{1}{120}+\frac{2}{2003}\)

\(\Rightarrow A=\frac{1}{180}+\frac{2}{2003}\)

\(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)

\(\Rightarrow B=\frac{5}{4}\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2003}\)

\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2003}\)

\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2003}\)

\(\Rightarrow B=\frac{5}{4}.\frac{1}{80}+\frac{5}{2003}\)

\(\Rightarrow B=\frac{1}{64}+\frac{5}{2003}\)

\(\left\{\begin{matrix}\frac{1}{64}>\frac{1}{180}\\\frac{5}{2003}>\frac{2}{2003}\end{matrix}\right.\Rightarrow\frac{1}{64}+\frac{5}{2003}>\frac{1}{180}+\frac{2}{2003}\Rightarrow B>A\)

Vậy A < B

20 tháng 4 2019

\(A=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}+\frac{2}{2003}\)

\(\text{Đặt }C=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}\)

\(C=\frac{2}{3}\left(\frac{3}{60\cdot63}+\frac{3}{63\cdot66}+...+\frac{3}{117\cdot120}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)\)

\(C=\frac{2}{3}\cdot\frac{1}{120}\)

\(C=\frac{1}{180}\)

\(\text{Thay }C=\frac{1}{180}\text{Ta có : }\) \(A=\frac{1}{180}+\frac{2}{2003}\)

\(B=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}+\frac{5}{2003}\)

\(\text{Đặt }D=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+...+\frac{5}{76\cdot80}\)

\(D=\frac{5}{4}\left(\frac{4}{40\cdot44}+\frac{4}{44\cdot48}+...+\frac{4}{76\cdot80}\right)\)

\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)\)

\(D=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)\)

\(D=\frac{5}{4}\cdot\frac{1}{80}\)

\(D=\frac{1}{64}\)

\(\text{Thay }D=\frac{1}{64}\text{ Ta có : }B=\frac{1}{64}+\frac{5}{2003}\)

\(\text{Vì }A=\frac{1}{180}+\frac{2}{2003}\text{ , }B=\frac{1}{64}+\frac{5}{2003}\)

\(\text{Có : }\frac{1}{180}< \frac{1}{64}\)

\(\frac{2}{2003}< \frac{5}{2003}\)

\(\Rightarrow\text{ }A< B\)

14 tháng 7 2018

A=2/3.(3/60.63+3/63.66+.....+3/117.120+3/120.123)

A=2/3.(1/60-1/63+1/63-1/66+...+1/117-1/120+1/20-1/123)

A=2/3.(1/60-1/123)

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