cmr neu 1/a+1/b+1/c =1/a+b+c thi 1/a^99+1/b^99+1/c^99=1/a^99+b^99+c^99

(1+97)*(1+97/2)*(1+97/3)*(1+97/4)... ... ...(1+97/99) 

(1+99)*(1+99/2)*(1+99/3)*(1+99/4)... ... ...(1+99/97) 

\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}\)=0

1+1+1+1+1+2+2+2+2+2+......+99+99+99+99+99+100+100+100+100+100=?

tinh tong gia tri bieu thuc :

a)A=1+1/3+1/5+...+1/97+1/99/1/1*99+1/3*97+1/5*95+...+1/97*3+1/99*1

b)B=1/2+1/3+1/4+...+1/100/99/1+98/2+97/3+...+1/99

Cho S = 99⁹⁹!+( 99⁹⁹+1)!/1!+( 99⁹⁹+2)!/2!+…+( 99⁹⁹+1000)!/1000!

Viết M = (99^99+1).S dưới dạng thương của hai giai thừa

1+1+1+1+2+2+2+2+3+3+3+3+...+99+99+99+99+100+100+100+100=?

E=(1+1/3+1/5+........+1/97+1/99)/(1/1*99+1/3*97+1/5*95+.....+1/97*3+1/99*1)

\(H=\frac{\left(1+97\right)\left(1+\frac{97}{2}\right)\left(1+\frac{97}{3}\right)\left(1+\frac{97}{4}\right)+...+\left(1+\frac{97}{99}\right)}{\left(1+99\right)\left(1+\frac{99}{2}\right)\left(1+\frac{99}{3}\right)\left(1+\frac{99}{4}\right)+...+\left(1+\frac{99}{97}\right)}\)

\(\frac{99^1}{1}+\frac{99^2}{1}+\frac{99^3}{1}+...+\frac{99^{100}}{1}\)so sánh với 100^{1 vạn}