A = 2/1*5 + 2/5*9 + ... + 2/101*105

   = 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)

   = 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)

   = 1/2(1 - 1/105)

   = 1/2 * 104/105 = 52/105

Sửa câu b. Phân số thứ 2 phải là 4/5*8

B = 4/2*5 + 4/5*8 + ... + 4/47*50

   = 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)

   = 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)

   = 4/3(1/2 - 1/50)

   = 4/3 * 24/50 = 16/25

a: =91/105+60/105-101/105

=50/105=10/21

c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)

d: =2-2/9

=18/9-2/9

=16/9

e: =24/36-9/36+8/36

=23/36

g: =5/2+1/2

=3

Ta có 1/2*3=1/2-1/3;

         1/3*4=1/3-1/4

       ......................(tương tự với các số khác)

         1/149*150=1/149-1/150

=>A=1/2-1/3+1/3-1/4+1/4-1/5+...-1/149+1/149-1/150=1/2-1/150

A=75/150-1/150=74/150=37/75

Vậy A= 37/75

A=1/2-1/3+1/3-1/4+...+1/149-1/150

=1/2-1/150

=37/75

B=3A=37/25

\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{149\cdot150}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{149}-\frac{1}{150}\)

\(A=\frac{1}{2}-\frac{1}{150}\)

\(A=\frac{75}{150}-\frac{1}{150}\)

\(A=\frac{74}{150}=\frac{37}{75}\)

Vậy \(A=\frac{37}{75}\)

c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)

\(=\dfrac{255}{103}\)

a)\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}\) 

=\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-2}{5}+\dfrac{-3}{5}\)

=\(-\dfrac{2}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{-3}{5}\) 

=\(\dfrac{-2}{5}.1+\dfrac{-3}{5}\) 

=\(-\dfrac{2}{5}+\dfrac{-3}{5}\) 

=\(-\dfrac{5}{5}\) = -1

\(\dfrac{5}{9}.\dfrac{14}{17}+\dfrac{1}{17}.\dfrac{5}{9}+\dfrac{2}{9}+\dfrac{5}{12}\)

=\(\dfrac{5}{9}.\left(\dfrac{14}{17}+\dfrac{1}{17}\right)+\dfrac{2}{9}+\dfrac{5}{12}\) 

=\(\dfrac{5}{9}.\dfrac{15}{17}+\dfrac{2}{9}+\dfrac{5}{12}\) 

=\(\dfrac{25}{51}+\dfrac{2}{9}+\dfrac{5}{12}\)

=\(\dfrac{691}{612}\)

\(-\dfrac{4}{25}:\dfrac{11}{79}+\dfrac{15}{11}:\dfrac{25}{79}\)

=\(-\dfrac{4}{25}.\dfrac{79}{11}+\dfrac{15}{11}.\dfrac{79}{25}\)

=\(-\dfrac{4}{25}.\dfrac{79}{11}+\dfrac{15}{25}.\dfrac{79}{11}\) 

=\(\left(-\dfrac{4}{25}+\dfrac{15}{25}\right).\dfrac{79}{11}\)

=\(\dfrac{11}{25}.\dfrac{79}{11}\) 

=\(\dfrac{79}{25}\)

a)\(1-2+3-4+5-6+7-8+8-9+9-10\)

=\(\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+\left(8-9\right)+\left(9-10\right)\)

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

\(=\left(-1\right).6\)

\(=-6\)

b)\(1-2+3-4+...+99-100\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)}\(\left[\left(100-1\right):1+1\right]:2=50\)(cặp)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)} 50 số (-1)

\(=\left(-1\right).50\)

\(=-50\)

c)\(1-3+5-7+9-11+13-15\)

\(=\left(1-3\right)+\left(5-7\right)+\left(9-11\right)+\left(13-15\right)\)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)

\(=\left(-2\right).4\)

\(=-8\)

d)\(1-3+5-7+...-99+101\) (Đối với bài này, có vẻ đề sai, mình đã sửa lại rồi

\(=\left(1-3\right)+\left(5-7\right)+...+\left(97-99\right)+101\) } \(\left[\left(99-1\right):2+1\right]:2=25\)(cặp)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\) } 25 số (-2)

\(=\left(-2\right).25\)

\(=-50\)

e)\(-1-2-3-4-...-99-100\)

\(=\left(-1\right)+\left(-2\right)+\left(-3\right)+...+\left(-99\right)+\left(-100\right)\)

\(=\left[\left(-1\right)+\left(-100\right)\right]+\left[\left(-2\right)+\left(-99\right)\right]+...+\left[\left(-51\right)+\left(-50\right)\right]\) } \(\left[\left(100-1\right):1+1\right]:2=50\)(cặp) (phần này của đề bài, không thay được như (-100) hoặc (-1))

\(=\left(-100\right)+\left(-100\right)+\left(-100\right)+...+\left(-100\right)\)} 50 số (-100)

\(=\left(-100\right).50\)

\(=-5000\)

a, -5

b, -50

c, -8

d, -50

e, -5050

cách làm nữa nha