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Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )
Đặt A = ( 1/1.2 + 1/3.4 + ... + 1/2017.2018)
= 1 - 1/2 + 1/3 - 1/4 + ... + 1/2017 - 1/2018
= ( 1 + 1/3 + ... + 1/2017 ) - ( 1/2 + 1/4 + ... + 1/2018 )
= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )
= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )
= 1/1010 + 1/1011 + ... + 1/2018
=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018
=> S = 1/2018
Vậy S = 1/2018
Ta có: \(\left(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\right)\div\frac{99}{100}\)
\(=2017\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\cdot\frac{100}{99}\)
\(=2017\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot\frac{100}{99}\)
\(=2017\cdot\left(1-\frac{1}{100}\right)\cdot\frac{100}{99}\)
\(=2017\cdot\frac{99}{100}\cdot\frac{100}{99}\)
\(=2017\)
Ta có\(\left(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\right):\frac{99}{100}\)
Đặt B=\(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\)
Ta có B =\(2017.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=2017.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2017.\left(1-\frac{1}{100}\right)=2017.\frac{99}{100}\)
Thay B vào A ta có A=\(2017.\frac{99}{100}:\frac{99}{100}=2017\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
\(\Rightarrow\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
co m/n =2017/2017 => m/n=1 =>m=n => m+2017=n+2017
suy ra m+2017/n+2017 =1
ma m/n=1 => m/n=m+2017/n+2017
Ta có :
\(\frac{m}{n}=\frac{2017}{2017}\Leftrightarrow m=n\)
=> \(\frac{m+2017}{n+2017}=\frac{m+2017}{m+2017}=1=\frac{m}{n}\)
=> \(\frac{m}{n}=\frac{m+2017}{n+2017}\)(đpcm)
a) ta có: \(1-\frac{2016}{2017}=\frac{1}{2017}\)
\(1-\frac{2017}{2018}=\frac{1}{2018}\)
\(\Rightarrow\frac{1}{2017}>\frac{1}{2018}\Rightarrow1-\frac{2016}{2017}>1-\frac{2017}{2018}\Rightarrow\frac{2016}{2017}< \frac{2017}{2018}\)
b) ta có: \(\frac{2017}{2016}-1=\frac{1}{2016};\frac{2018}{2017}-1=\frac{1}{2017}\)
\(\Rightarrow\frac{1}{2016}>\frac{1}{2017}\Rightarrow\frac{2017}{2016}-1>\frac{2018}{2017}-1\Rightarrow\frac{2017}{2016}>\frac{2018}{2017}\)
\(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2017.\left(1-\frac{1}{100}\right)\)
\(=2017.\frac{99}{100}\)
\(=\frac{199693}{100}\)
\(\frac{2017}{1.2}+\frac{2017}{3.4}+\frac{2017}{4.5}+...+\frac{2017}{99.100}\) \(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\) \(=2017.\left(1-\frac{1}{100}\right)\) \(=2017.\frac{99}{100}\) \(=\frac{199693}{100}\)