\(I=\left(1+5+...+2009\right)+\left(2+6+...+2010\right)-\left(3+7+...+2007\right)-\left(4+8+...+2008\right)\)

\(=\frac{\left(2009+1\right).\left[\left(2009-1\right):4+1\right]}{2}+\frac{\left(2010+2\right)\left[\left(2010-2\right):4+1\right]}{2}\)

\(-\frac{\left(2007+3\right).\left[\left(2007-3\right):4+1\right]}{2}+\frac{\left(2008+4\right)\left[\left(2008-4\right):4+1\right]}{2}\)

\(=\frac{2010.503}{2}+\frac{2012.503}{2}-\frac{2010.502}{2}-\frac{2012.502}{2}\)

\(=1005+1006=2011\)

 \(A=1+3-4+11-15+19-23+...+99-103\)

\(A=1+\left(3-7\right)+\left(11-15\right)+\left(19-23\right)+...+\left(99-103\right)\)

\(A=1+\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(A=1+\left(-4\right).13\)

\(A=1+\left(-52\right)\)

\(A=-51\)

nhớ là giải rõ nha

dễ ẹc

Ta có :

3-7=11-15=...=99-103

Từ 3 đến 103 có số số hạng là :

( 103 - 3 ) : 4 + 1 = 26 ( số )

Từ 3 đến 103 có số cặp là :

26 : 2 = 13 ( cặp )

Vậy1+ 3-7+11-15+19-23+...+99-103 = 13 . (-4) +1 = -51

Đáp Số : -51 

    (1 + 2 + 3 + 4 + ... + 2010).(1 + 2² + 3³ + ... + 2010²⁰¹⁰ + 2011²⁰¹¹).(170170 - 7.11.13.170)

= (1 + 2 + 3 + 4 + ... + 2010).(1 + 2² + 3³ + ... + 2010²⁰¹⁰ + 2011²⁰¹¹).(170170 - 170170)

= (1 + 2 + 3 + 4 + ... + 2010).(1 + 2² + 3³ + ... + 2010²⁰¹⁰ + 2011²⁰¹¹).   0

=                     0

A.5/7

B.-481/391

C.0

D.-59/9

E.1

\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)

=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)

= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)

= \(1\dfrac{11}{25}+\dfrac{5}{7}\)

= \(2\dfrac{27}{175}\)

b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)

=\(-2+\left(\dfrac{ }{ }\right)\)

Số số hạng của  tổng B là :

(103-3): 4 +1 = 26 ( số hạng )

Tổng B là :

(103+3).26:2 = 1378 

số số hạng là : (103-3)/4 +1= 26 ( số hạng )

Tổng B là : (103+3).26/2= 1378

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

Các bạn giải giúp mình nha😐

hơi lệch

\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)

\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)

Bài 1: 

Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)

Bài 2: 

Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)

\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=7\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{63}{10}\)

a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

      \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

       \(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

                  \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

                   \(=3.\left(1-\frac{1}{100}\right)\)

                     \(=3.\frac{99}{100}=\frac{297}{100}\)