1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)

   A= \(\frac{100}{2}\)

   A=50

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

     B= \(\frac{1}{99}\)

\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)

     \(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)

     \(=\frac{100}{2}\)

       \(=50\)

\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)

     \(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)

       \(=-\frac{1}{99}\)

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)..........\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)

\(=\frac{100}{2}=50\)

\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)

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

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

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)

  \(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5.....100}{2.3.4.....99}\)

 \(=\frac{100}{2}=50\)

ai trả lời đúng k

có cách làm nữa nha

a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)

a) =3/2 . 4/3 . 5/4 ...100/99

   =\(\frac{3.4.5...100}{2.3.4..99}\)

  =\(\frac{100}{2}\)

b) =

b) = \(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=-1\left(\frac{1.2.3...99}{2.3.4...100}\right)=-\frac{1}{100}\)

21=45

1+1=3 :)))

\(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):....:\left(\frac{1}{100}-1\right)\text{ có số số lẻ thừa số âm nên bằng:}\)

\(-\left[\left(1-\frac{1}{2}\right):\left(1-\frac{1}{3}\right):...\left(1-\frac{1}{100}\right)\right]=-\left[\frac{1}{2}:\frac{2}{3}:\frac{3}{4}:......:\frac{99}{100}\right]=-\left(\frac{1.3.4...100}{2.2.3...99}\right)=-50\)

bn thử nghĩ xem

mk chắc chắn bn lm được]

Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)

\(\Rightarrow A=1-\frac{1}{2^{99}}=\frac{2^{99}-1}{2^{99}}\)

A=50/99