1/1x2 + 1/2x3 + 1/3x4 + ... 1/99 x 100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
haizzz đáng tiếc tôi muốn ns là: ko bao f và đừng mong chờ OK
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
Lên Qanda mà hỏi
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
tính tổng: 1/1x2 + 1/2x3 + 1/ 3x4 +...+ 1/98 x 99 + 1/99 x 100
1/1*2 + 1/2*3 + 1/3*4 + .... + 1/99 * 100
= 1- 1/100
= 99/100
=1-1/2+1/2-...-1/99+1/99-1/100=1-1/100=99/100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
ai lam dc mik tick cho
Tìm x biết 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/X(X+1) = 99/100
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99