mk gợi ý thui nhé :

cộng 96 phân số theo từng cặp:

a/b = (1/1+1/96)+(1/2+1/95)+(1/3+1/94)+...+(1/48+1/49)

...........................v.v

tự làm nhé

cho mk xin cái k

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

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

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

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

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

Vậy...

P/s: Hoq chắc

#)Giải :

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

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

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

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

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

Ta có:

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

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

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

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

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

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

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

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

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

(số hạng 99 chia thảnh 99 số 1 cộng vào từng phân số còn dư 1 số 1 để ngoài)

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

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

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

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

\(\Rightarrow\frac{B}{A}=100\)

b/a = 100. Nếu k đúng cho mình, Mình sẽ trình bày cách làm cho bạn.

ui hình như đây là đề kiểm tra toán cuối năm học lớp 6 đúng ko zậy???

ryrd6s4rtg

A : B = 2019

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

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

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

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

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