C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

  =\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

  =\(\frac{1}{100}-\frac{99}{100}\)

  =\(\frac{-98}{100}=\frac{-49}{50}\)

C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1 
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1) 
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A 
Dễ thấy 1/2.1 = 1/1 - 1/2 
1/3.2 = 1/2 - 1/3 
..................... 
1/99.98 = 1/98 - 1/99 
1/100.99 = 1/99 - 1/100 
=> cộng từng vế với vế ta

\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{-5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)\)

\(+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=0+0+0+0-\frac{1}{16}\)

\(=\frac{-1}{16}\)

       A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101

=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4

=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)

=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101

=> 4A = 99*100*101*102

=> 4A = 101989800

=>   A = 25497450

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

T=(12−13)(12−15)(12−17).....(12−199)T=(12−13)(12−15)(12−17).....(12−199)

⟹T=12(1−23).12(1−25).12(1−27).....12(1−299)⟹T=12(1−23).12(1−25).12(1−27).....12(1−299)

Thấy T có: (99-3):2+1=49(SH)

⟹T=(12.49).[(1−23).(1−25)...(1−299)⟹T=(12.49).[(1−23).(1−25)...(1−299)

⟹T=492.199=49198

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

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

     \(=1-\frac{1}{100}=\frac{99}{100}\)

a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)

\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)