a,A=1 + ( -3) + 5 + ( -7 ) + ... + 17 + ( -19 )

A=( 1 - 3 ) + ( 5 - 7 ) + ...+ ( 17 +19 )

A= (-2 ) . 10

A= (-20)

b, B= 1-4+7-10 +... -100 + 103

B= 1+ ( -4 + 7 ) + ( -10 +13 ) +...+ (-100 +103 )

B= 1 + 3 + 3 +...+3

B= 1+3 .17

B= 52

c, C= 1 + 2 -3 -4+5+6-7-8+..-99-100+101+102

C= 1 + ( 2-3-4+5) +(6-7 -8+9)+...+(98-99-100+101)+102

C= 1 + 0 + 0 + 0 + 0 + ... + 0 + 102

C= 103

A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 99 - 100 + 101 + 102

A=(1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) + ... (97 + 98 - 99 - 100) + 101 + 102

A=(-4) + (-4) +...+ (-4) + 203 ( có 25 số -4)

A=25.(-4)+203

A=-100+203

A=103

B = 1 + (-3) + 5 + (-7) + …+ 17 + (- 19)  

B=[1 + (-3)] + [5  +(-7)] +...+ [17 + (-19)]          Có 5 cặp số

B=(-2) + (-2) +...+ (-2)             có 5 số hạng

B=(-2).5

B=-10

C = 1 -  4 + 7 - 10 + … - 100 + 103

C = (1 -  4) + (7 - 10) + … +(97- 100) + 103          có 34 cặp số

C=(-3) + (-3) +...+ (-3)  +103           có 34 số -3

C=34.(-3)+103

C=-102+103

C=1

A = 103

A =103

B=-10

C=1

Nói chung là làm như bạn The love of Shinichi and Ran immortal so no one can change it bạn ý làm đúng rồi đấy!!

~chúc bạn học giỏi~

A=1+(-3)+5+(-7)+...17+(-19)

=> A=(1+5+9+13+17)-(3+7+11+15+19)

=>A=45-55

=>A=-10

Ta có : 

A=1+(-3)+5+(-7)+...17+(-19)

=> A=(1+5+9+13+17)-(3+7+11+15+19)

=>A=45-55

=>A=-10

Đap số : -10

\(a,\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{9}=\dfrac{12}{15}+\dfrac{10}{15}+\dfrac{1}{9}=\dfrac{22}{15}+\dfrac{1}{9}=\dfrac{66}{45}+\dfrac{5}{45}=\dfrac{71}{45}\)

\(b,\dfrac{3}{7}+\dfrac{11}{14}+\dfrac{19}{28}=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}=\dfrac{53}{28}\)

\(c,\dfrac{1}{2}+\dfrac{1}{7}+\dfrac{-1}{5}=\dfrac{7}{14}+\dfrac{2}{14}+\dfrac{-1}{5}=\dfrac{9}{14}+\dfrac{-1}{5}=\dfrac{45}{70}+\dfrac{-14}{70}=\dfrac{31}{70}\)

\(d,\dfrac{7}{8}+\dfrac{5}{16}+\dfrac{-3}{4}=\dfrac{14}{16}+\dfrac{5}{16}+\dfrac{-12}{16}=\dfrac{7}{16}\)

\(e,\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{3}{12}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{8}{12}+\dfrac{-1}{13}=\dfrac{2}{3}+\dfrac{-1}{13}=\dfrac{26}{39}+\dfrac{-3}{39}=\dfrac{23}{39}\)

\(g,\dfrac{2}{3}+\dfrac{3}{8}+\dfrac{-5}{12}=\dfrac{16}{24}+\dfrac{9}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-10}{24}=\dfrac{15}{24}\)

de ma

ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều 

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)