Question

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

Answer 1

Có : (1+1/2+1/3+....+1/100)+(1/2+2/3+....+99/100)

= 1+(1/2+1/2)+(1/3+2/3)+.....+(1/100+99/100) ( có 99 cặp )

= 1+1+1+....+1 ( có 100 số 1 )

= 100

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

Tk mk nha

Answer 2

vì sao đang bằng lại chuyển thành cộng

Answer 3

Vì theo quy tắc chuyển vế ta có :

a - b = c thì a = b+c

Tk mk đi

Answer 4

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

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

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

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

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