42x13-22x5+42x7-15x22

= 42 x (13+7) - 22 x (5+15)

= 42 x 20 - 22 x 20

= 20 x (42 - 22)

= 20 x 20 = 400

___

37x39+78x74+13x85+52x55

= 37 x 3 x 13 + 13 x 6 x 74 + 13 x 85 + 13 x 4 x 55

= 13 x (37 x 3+ 74 x 6+ 85 + 4 x 55)

= 13 x (111 + 444 + 85 + 220)

= 13 x 860 = 11180

(637x527-189):(526x637+448)

= (637 x 526 + 637 - 189): (526 x 637 + 448)

= (637 x 526 + 448) : (526 x 637 + 448) = 1

\(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{52.55}+\frac{1}{55.58}\)

\(\Rightarrow C=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{52.55}+\frac{3}{55.58}\right)\)

\(\Rightarrow C=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{52}-\frac{1}{55}+\frac{1}{55}-\frac{1}{58}\right)\)

\(\Rightarrow C=\frac{1}{3}.\left(1-\frac{1}{58}\right)\)

\(\Rightarrow C=\frac{1}{3}.\frac{57}{58}\)

\(\Rightarrow C=\frac{19}{58}\)

Vậy \(C=\frac{19}{58}\)

~ Ủng hộ nhé 

\(3C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{1}{55\times58}..\)

\(3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}....+\frac{1}{55}-\frac{1}{58}..\)

\(3C=1-\frac{1}{58}=\frac{57}{58}\)

\(C=\frac{57}{174}\)

Ai tên triết đừng xem nhé ahihi

Ta có : A = 1.3 + 3.5 + 5.7 + ....... + 37.39 + 39.41

=> 6A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ....... + 39.41.43

=> 6A = 39.41.43

=> A = 39.41.43 / 6

=> A = 11459,5

A=1x3+3x5+5x7+...+37x39+39x41

6A=1x3x5-1x3x5+3x5x7-3x5x7+...+39x41x43

6A=39x41x43

A=\(\frac{39x41x43}{6}\)

\(A=\dfrac{1}{3.5}+\dfrac{1}{7.9}+...+\dfrac{1}{37.39}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}.\dfrac{4}{13}\\ =\dfrac{2}{13}\)

A=13.5+17.9+...+137.39=12(23.5+27.9+...+237.39)=12(13−15+15−17+...+137−139)=12(13−139)=12.413=213

a) \(a=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)

\(a=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(a=\frac{1}{3}-\frac{1}{39}\)

\(a=\frac{12}{39}\)

b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-2}{12}+\frac{2}{12}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot0\)

\(=0\)

a) \(A=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{37x39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{4}{13}\)

b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x0\)

\(=0\)

B= 2/3.5 + 2/5.7 +.....+ 2/37.39

  = 1/3 - 1/5 + 1/5 - 1/7 + ...... + 1/37 - 1/39

  = 1/3 - 1/39 =12/39

C = 1/4 - 1/7 + 1/7 - 1/10 + .....+ 1/73 - 1/ 76

  = 1/4 - 1/76 = 18/76

B= 2/3.5 + 2/5.7 +.....+ 2/37.39

  = 1/3 - 1/5 + 1/5 - 1/7 + ...... + 1/37 - 1/39

  = 1/3 - 1/39 =12/39

C = 1/4 - 1/7 + 1/7 - 1/10 + .....+ 1/73 - 1/ 76

  = 1/4 - 1/76 = 18/76

1/2 x (6/1-6/3+6/3-6/5+ ... +6/37-6/39)

1/2 x (6/1-6/39)

1/2 x 228/39

228/78

\(\frac{6}{1\cdot3}+\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{35.37}+\frac{6}{37.39}\) 

\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{39}\right)\) 

\(=3\left(1-\frac{1}{39}\right)=3\cdot\frac{38}{39}=\frac{114}{39}\) 

Ps: Bạn tự rút gọn nhé!!!

\(\frac{6}{1.3}+\frac{6}{3.5}+...+\frac{6}{35.37}+\frac{6}{37.39}\)

\(=3\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{35.37}+\frac{2}{37.39}\right)\)

\(=3\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{39}\right)\)

\(=3\left(1-\frac{1}{39}\right)\)

\(=3.\frac{38}{39}\)

\(=\frac{38}{13}\)

Bài 2 :

Ta có :

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)\)

    \(=1-\frac{1}{32}\)

và \(B=\frac{2003}{2004}=1-\frac{1}{2004}\)

Vì \(\frac{1}{32}>\frac{1}{2004}\) nên A < B

A<B