\(\frac{1.4}{2.3}\)x\(\frac{2.5}{3.4}\)x\(\frac{3.6}{4.5}\)x.........x\(\frac{2013.2016}{2014.2015}\)=\(\frac{1.2.3....2013}{2.3.4...2014}\)x \(\frac{4.5.6....2016}{3.4.5....2015}\)

                                                                                   =\(\frac{1}{2014}\)x \(\frac{2016}{3}\)

                                                                                   =\(\frac{2016}{6042}\)= \(\frac{336}{1007}\)

giải chưa nhở

mk cx đg cần giải bài này

\(A=\frac{4}{6}+\frac{10}{12}+\frac{18}{20}+...+\frac{9898}{9900}\)

\(A=1-\frac{2}{6}+1-\frac{2}{12}+1-\frac{2}{20}+...+1-\frac{2}{9900}\)

\(A=98-\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\right)\)Đặt Biểu thức trong ngoặc đơn là B

\(\Rightarrow A=98-B\)

\(\Rightarrow\frac{B}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(\frac{B}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)

\(\frac{B}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Rightarrow B=\frac{2.49}{100}=\frac{98}{100}\)

Ta nhận thấy \(B=\frac{98}{100}< 1\Rightarrow A=98-\frac{98}{100}=97+\frac{2}{100}\)

\(\Rightarrow97< A< 98\left(dpcm\right)\)

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)

E=\(\frac{1.2.3.....97.98}{2.3.4.....98.99}\)+\(\frac{4.5.6....100.101}{3.4.5...99.100}\)

E=\(\frac{1}{99}\)+\(\frac{101}{3}\)

E=\(\frac{304}{99}\)

N=\(\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+....+\)\(\frac{98.101}{99.100}\)

N=\(\frac{1.2.3...98}{2.3.4...99}\)\(+\)\(\frac{4.5.6....101}{3.4.5....100}\)

N=\(\frac{1}{99}+\frac{101}{3}\)

N=\(\frac{3334}{99}\)