Bài 1: 

\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)

\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)

\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)

=1/131

Bài 2:

b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)

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

\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)

b) \(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100.99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(=\dfrac{1}{100.99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{100.99}-\left(1-\dfrac{1}{99}\right)\)

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

1/2+2/5+1/3+5/7+1/6+(-4/35)+1/41

đổi ra rồi nha bạn thế là tự biết làm được rồi

     A = 1/99 - 1/99.98 - 1/98.97 - ............... - 1/3.2 - 1/2.1

\(A=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(B=\frac{1}{99.98}+\frac{1}{97.87}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

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

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

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

\(B=\frac{98}{99}\)

\(\Rightarrow A=\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

c/
C = 1/100-1/100-1/99-1/99-1/98-1/98-1/97-..........-1/3-1/2-1/2-1/1
C = 1/100-1/100-1/1
C = 0-1/1
C = -1

1/99-1/99.98-1/98.97-.....-1/2.1

= 1/99-(1/99.98+1/98.97+......+1/2.1

=1/99-(1/99-1)

=1

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

-49/50