Ta có 99/1+98/2+97/3+...+1/99=(98/2+1)+(97/3+1)+...+(1/99+1)+1

=100/2+100/3+...+100/99+100/100

=100(1/2+1/3=1/4+1/5+...+1/99+1/100)

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

xét mẫu số = \(\frac{99}{1}\)+\(\frac{98}{2}\)+....+\(\frac{1}{99}\)

mẫu số = (\(1+\frac{98}{2}\))+(\(1+\frac{97}{3}\))+.......+(\(1+\frac{1}{99}\))

mẫu số = \(\frac{100}{2}\)+\(\frac{100}{3}\)+....+\(\frac{100}{99}\)

mẫu số =100 x (\(\frac{1}{2}\)+\(\frac{1}{3}\)+....+\(\frac{1}{99}\))             (1)

thay (1) vào biểu thức trên

1/2+1/3+1/4+.....+1/100  /   100 x (1/2+1/3+...+1/99)

= \(\frac{1}{100}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1}\)

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

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

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

M=100

Xét tử N

92-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)

=(1+1+1+...+1)-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)

=1-(1/9)+1-(2/10)+1-(3/11)+......+1-(90/98)+1-(91/99)+1-(92/100)

=(8/9)+(8/10)+(8/11)+ ...+ (8/98)+(8/99)+(8/100)

=8.[(1/9)+(1/10)+(1/11)+...+(1/98)+(1/99)+(1/100)]

=40[(1/45)+(1/50)+(1/55)+...+(1/495)+(1/500)]

=>N=40

=>M/N=5/2

1) \(+2x+3y⋮17\)

\(\Rightarrow26x+39y⋮17\)

\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)

Mà \(17x+34y⋮17\)

\(\Rightarrow9x+5y⋮17\)

\(+9x+5y⋮17\)

\(\Rightarrow36x+20y⋮17\)

\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)

Mà \(34x+17y⋮17\)

\(\Rightarrow2x+3y⋮17\)