a,=(1/3+3/5+1/15)+(3/4+-1/36)+(1/72-2/9)=1+26/36-15/72=1+(52-15)/72=1+37/72=109/72

b,=1/100-(1/1x2+1/2x3+...+1/97x98+1/98x99+1/99x100)

   =1/100-(1/1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)

   =1/100-(1/1-1/100)=1/100-99/100=-98/100=-49/50

chỉ có mk mk giải thôi đó l-i-k-e đi

mình cũng đang bí bài này

1/2!+1/3!+...+1/100!<1/1*2+1/2*3+1/3*4+...+1/99*100

1-1/100<1

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

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

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

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

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

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

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

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

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

      \(=1\)

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

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)

Toán lớp 9 à ? Năm nay hết hè mk mới lên Lớp 6 nên mk k thể làm đc vì mk chưa học .Chúc bn sẽ sớm tìm đc đáp án đúq ^.^

Bài làm:

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

\(=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{99.100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}\right)\)

\(=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{99.100}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{99.100}-\frac{98}{99}\)

\(=\frac{1-98.100}{99.100}=\frac{1-9800}{9900}=-\frac{9799}{9900}\)

Học tốt!!!!

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

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

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

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

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

\(=\frac{2}{99}-\frac{101}{100}\)

-49

mình giải rồi vòng 11 đó 

Chọn C