Question

Tính Q=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}{\frac{100-1}{1}+\frac{102-2}{2}+...+\frac{100-99}{99}}\)

Answer 1

Sửa đề:

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

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

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

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

Answer 2

haha!dungs rois!

Answer 3

sai đề bài kìa

Answer 4

trả lời:

1/100 nhé bạn

      hok tốt.

Answer 5

Đáp án:\(\frac{1}{100}\)

Answer 6

trả lời: \(\frac{1}{100}\) nha

😁 😁 😁

Answer 7

Xét K=\(\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}\)

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

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

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

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

=> Q=1/100