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\(\frac{1.4}{2.3}\)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 \(\frac{2016}{3}\)
=\(\frac{2016}{6042}\)= \(\frac{336}{1007}\)
Lời giải:
$M=\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+...+\frac{98.101}{99.100}$
$=1-\frac{2}{2.3}+1-\frac{2}{3.4}+1-\frac{2}{4.5}+...+1-\frac{2}{99.100}$
$=(1+1+....+1)-2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100})$
$=98-2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100})$
$=98-2(\frac{1}{2}-\frac{1}{100})$
$=97+\frac{1}{50}=97,02$
\(A=\frac{4}{6}+\frac{10}{12}+\frac{18}{20}+...+\frac{9898}{9900}\)
\(A=1-\frac{2}{6}+1-\frac{2}{12}+1-\frac{2}{20}+...+1-\frac{2}{9900}\)
\(A=98-\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\right)\)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=\(\frac{3334}{99}\)
\(C=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{2013.2016}{2014.2015}\)
\(C=\frac{\left(1.2.3.....2013\right).\left(4.5.6.....2016\right)}{\left(2.3.4.....2014\right).\left(3.4.5.....2015\right)}\)
\(C=\frac{1}{2014}.\frac{2016}{3}\)
\(C=\frac{336}{1007}\)