Xin hãy giúp mình

a,=3/2*4/3*....100/99

=3*4*5*....*100/2*3*...*99

=100/2=50

b, nhân lên băng:

1*2*3*...*99/2*3*...*100=1/100

câu 1 :1/100

câu 2 :1

b)ta đặt A:  \(A=\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\)

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

                  \(A=\frac{100}{99}+\frac{100}{98}+..+\frac{100}{2}+\frac{100}{100}\)

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

a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)

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

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

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

\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)

b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)

\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)

\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)

Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)

c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)

\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)

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

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

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

Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)​

\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)

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

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

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

\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)

\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)

Vậy \(A:B=1.\)

b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)

\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)

\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)

\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)

Vậy \(A:B=1.\)

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

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

\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)

\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)

\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)

Vậy \(A:B=1.\)

1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+...+1/98*99+1/99*100

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{50}{100}-\frac{1}{100}\)

\(=\frac{49}{100}\)

1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ....... + 1/98 x 99 + 1/99 x 100

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ..... + 1/98 - 1/99 + 1/99 - 1/100

= 1/2 - 1/100

= 49/100

đề bảo tính gì vậy?

\(\frac{1}{2}\div\frac{1}{3}\div\frac{1}{4}\div\frac{1}{5}\div...\div\frac{1}{98}\div\frac{1}{99}\div\frac{1}{100}\)

\(=\frac{1}{2}\cdot3\cdot4\cdot5\cdot...\cdot98\cdot99\cdot100\)

Phép tính như thế này tạm thời chuyển sang :

\(\frac{1}{2}\cdot\frac{100!}{2}=\frac{100!}{4}\)

= 24497550

Ta có:(1+2+3+4+5+...+97+98+99)x100/2x50:2500

        =[(99+1)x99:2)]x50x50:2500

       =4950x2500:2500

       =4950x1

      =4950