N = 1 - 2/2.3 + 1 - 2/3.4 +.....+ 1 - 2/99.100

   = 98 - 2.(1/2.3 + 1/3.4 + ...... + 1/99.100)

   = 98 - 2.(1/2-1/3+1/3-1/4+....+1/99-1/100)

   = 98 - 2.(1/2-1/100)

   = 98 - 2.49/100 = 98-49/50 < 98

Mà 49/50 < 1

=> N > 98-1 = 97

=> 97 < N < 98

Tk mk nha

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Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3

2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4

...

Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)

N=98+1/100−1/2−1/2.3−1/3.4−....−1/99.100

Xét P=1/2.3+1/3.4+....+1/99.100

P= 1/2−1/3+1/3−1/4+.....+1/99−1100 

P=1/2−1/100

Vậy N=98-1+1/50

N=97+1/50

Vậy 97<N<98(ĐPCM)

giải chưa nhở

mk cx đg cần giải bài này

Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3

2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4

...

Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)

N=\(98+\dfrac{1}{100}-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-....-\dfrac{1}{99.100}\)

Xét P=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{100}\)

Vậy N=98-1+\(\dfrac{1}{50}\)

N=\(97+\dfrac{1}{50}\)

Vậy 97<N<98(ĐPCM)

Ta có \(\frac{a\left(a+3\right)}{\left(a+1\right)\left(a+2\right)}=\frac{\left(a+1-1\right)\left(a+2+1\right)}{\left(a+1\right)\left(a+2\right)}=\frac{\left(a+1\right)\left(a+2\right)-\left(a+2\right)+\left(a+1\right)-1}{\left(a+1\right)\left(a+2\right)}\\ \)

= \(1-\frac{2}{\left(a+1\right)\left(a+2\right)}\)

Áp dụng ta có N = \(98-\left(\frac{2}{2.3}+...+\frac{2}{99.100}\right)=98-2.\left(\frac{1}{2.3}+...+\frac{1}{99.100}\right)=98-2.\left(\frac{1}{2}-\frac{1}{100}\right)>97\)