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

Làm theo cách này nhé : 

a = 2 / 60 x 63 + 2 / 63 x 66 + 2 / 66x 69 + ...+ 2 / 117 x 120 + 2 / 2011

= 2/3 x ( 3/60 x 63 +  3 / 63 x 66 + 3 / 66  x 69 + ...+ 3/117 x 120 ) + 2/2011

= 2/3 x ( 1/60 - 1/63 + 1/63 - 1/66 + 1/66 - 1/69 + ... + 1/117 - 1/120 ) + 2/2011

= 2/3 x  (  1/60 - 1/120 ) + 2/2011

= 2/3 x   1/120 + 2/2011

= 1/180 + 2/2011

b =  5/ 40 x 44 + 5 / 44 x 48 + ...+ 5/76 x 80 + 5/ 2011

= 5/4 x ( 4/40 x 44 + 4/44 x 48 + ...+ 4/76 x 80 ) + 5/2011

= 5/4  x ( 1/40 - 1/44 + 1/44 - 1/48 + ...+ 1/76 - 1/80 ) + 5/2011

= 5/4 x ( 1/40 - 1/80 ) + 5/2011

= 5/4 x        1/80 + 5/2011

= 1/64 + 5/2011

Do 1/64 > 1/80 ;   5/2011 > 2/2011

=> 1/64 + 5/2011 > 1/80 + 2/2011

=> b > a

K nha

Mình sửa lại chút nhé , lỗi đánh bàn phím thoy , :

Do 1/64 > 1/180 ;   5/2011 > 2/2011

=> 1/64 + 5/2011 > 1/180 + 2/2011

=> b > a 

b >a  bạn nhé

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)

+)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)

C1: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)

C2:

b: $A=$32​(601​−631​+631​−661​+...+1171​−1201​)+20032​

=23⋅1120+22003=32​⋅1201​+20032​

=2(1360+12003)=2(3601​+20031​)

�=54(140−144+144−148+...+176−180)+52003B=45​(401​−441​+441​−481​+...+761​−801​)+20035​

=54⋅180+52003=45​⋅801​+20035​

=5(1320+12003)=5(3201​+20031​)

Vì 1/360+1/2003<1/320+1/2003

nên A<B

vai dai

wa dai

b: �=23(160−163+163−166+...+1117−1120)+22003A=32​(601​−631​+631​−661​+...+1171​−1201​)+20032​

=23⋅1120+22003=32​⋅1201​+20032​

=2(1360+12003)=2(3601​+20031​)

�=54(140−144+144−148+...+176−180)+52003B=45​(401​−441​+441​−481​+...+761​−801​)+20035​

=54⋅180+52003=45​⋅801​+20035​

=5(1320+12003)=5(3201​+20031​)

Vì 1/360+1/2003<1/320+1/2003

nên A<B

300

300

= 300 

Ta có: \(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)

\(=\frac{124}{1984}\left(\frac{1984}{1.1985}+\frac{1984}{2.1986}+\frac{1984}{3.1987}+...+\frac{1984}{16.2000}\right)\)

\(=\frac{1}{16}\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+...+\frac{1}{16}-\frac{1}{2000}\right)\)

\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}+...+\frac{1}{2000}\right)\right]\)

\(B=\frac{1}{1.17}+\frac{1}{2.19}+...+\frac{1}{1984.2000}\)

\(=\frac{1}{16}\left(\frac{16}{1.17}+\frac{16}{2.18}+...+\frac{16}{1984.2000}\right)\)

\(=\frac{1}{16}\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)

\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)\right]-\left[\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right]\)

\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1984}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

Vậy A = B

dễ tự nghĩ

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

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

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

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

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

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

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

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

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

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

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

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

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

Vậy A < B