Ta có:

S=1.1/2+1/2.1/3+...+1/29.1/30

  =1-1/2+1/2-1/3+...+1/29-1/30

  =1-1/30=29/30

a) \(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

Đặt A = \(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)

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

A = \(1-\frac{1}{99}\)

A = \(\frac{98}{99}\)

Thay A vào ta được :

\(\frac{1}{100.99}-\frac{98}{99}=\frac{1}{9900}-\frac{98}{99}=\frac{-9799}{9900}\)

b) \(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-3,6.21\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

Ta thấy biểu thức trong ngoặc thứ ba của tử số có kết quả bằng 0

\(\Rightarrow\)Phân số ấy có kết quả bằng 0

\(a,\frac{1}{999\cdot1000}-\frac{1}{998\cdot999}-\frac{1}{997\cdot998}-...-\frac{1}{2\cdot1}\)

\(=\frac{1}{999\cdot1000}-\left[\frac{1}{2\cdot1}+\frac{1}{2\cdot3}+...+\frac{1}{997\cdot998}+\frac{1}{998\cdot999}\right]\)

\(=\frac{1}{999\cdot1000}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{998}-\frac{1}{999}\right]\)

\(=\frac{1}{999\cdot1000}-\left[1-\frac{1}{999}\right]=\frac{1}{999\cdot1000}-\frac{998}{999}=...\)

Tính nốt , không chắc :v

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

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

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

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

\(C=\frac{-98}{100}=\frac{-49}{50}\)

Ủng hộ mk nha ^_-

khó nhìn lắm bn ak

sao pn ko cho 

\(\frac{11}{125}-\frac{17}{18}-\frac{5}{8}+\frac{4}{9}+\frac{17}{14}.\)

thì có phải dễ nhìn hơn ko

a, \(A=\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)

\(=\frac{11}{125}+\left(\frac{-17}{18}+\frac{4}{9}\right)+\left(\frac{-5}{7}+\frac{17}{14}\right)\)

\(=\frac{11}{125}+\frac{-1}{2}+\frac{1}{2}\)

\(=\frac{11}{125}\)

b, \(B=1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)

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

\(=4-3=1\)

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

\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-49}{50}\)

a) Đặt biểu thức trên là A

\(A=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}+\frac{2}{11}-\frac{5}{7}+\frac{3}{7}-\frac{1}{5}\)

\(A=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{-2}{11}+\frac{2}{11}\right)+\frac{5}{9}+\frac{7}{13}-\frac{5}{7}\)

\(A=0+0+0+\frac{5}{9}+\frac{7}{13}-\frac{5}{7}\)

\(A=\frac{128}{117}-\frac{5}{7}\)

\(A=\frac{311}{819}\)

1. 1-2+3-4+5-6-.....+99-100

=(1-2)+(3-4)+(5-6)+...+(99-100)                              (50 cặp)

=(-1)+(-1)+(-1)+...+(-1)                                          (50 số -1)

=(-1).50

=-50

2.1+3-5-7+9+11-.....-397-399

=(1+3-5-7)+(9+11-13-15)+....+(387+389-391-393)+395-397-399 (99 cặp)

=(-8)+(-8)+(-8)+...+(-8)+(-401)(có 99 có -8)

=(-8).99+(-401)

=(-792)+(-401)

=-1193

3. 1-2-3+4+5-6-7+...+96+97-98-99+100

=(1-2-3+4)+(5-6-7+8)+...+(93-94-95+96)+(97-98-99+100)            (25 cặp)

=0+0+0+...+0

=0

4. A=2¹⁰⁰-2⁹⁹-2⁹⁸-.....-2²-2-1

2A=2¹⁰¹-2¹⁰⁰-2⁹⁹-....-2³-2²-2

2A-A=A=2¹⁰¹-2¹⁰⁰-2¹⁰⁰+1

A=2¹⁰¹-2.2¹⁰⁰+1

A=2¹⁰¹-2¹⁰¹+1

A=1