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\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
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ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
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theo bài ra ta có
\(\frac{a^{2015}}{b^{2017}+c^{2019}}=\frac{b^{2017}}{a^{2015}+c^{2019}}=\frac{c^{2019}}{a^{2015}+b^{2017}}\)
=>\(\frac{a^{2015}}{b^{2017}+c^{2019}}+1=\frac{b^{2017}}{a^{2015}+c^{2019}}+1=\frac{c^{2019}}{a^{2015}+b^{2017}}+1\)
=> \(\frac{a^{2015}+b^{2017}+c^{2019}}{b^{2017}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+b^{2017}}\)
- nếu a2015+ b2017 +c2019 = 0
=> b2017+ c2019 = -(a2015) (1)
=> a2015+ c2019= -(b2017) (2)
=> a2015+ b2017= -(c2019) (3)
thay 1, 2, 3 vào S ta có:
S = \(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}\)
=> S =\(\frac{-\left(a^{2015}\right)}{a^{2015}}+\frac{-\left(b^{2017}\right)}{b^{2017}}+\frac{-\left(c^{2019}\right)}{c^{2019}}\)
S = -1 + -1 + -1
S = -3
vậy S ko phụ thuộc vào giá trị a,b,c
- nếu a2015+b2017+c2019 khác 0
=> b2017+c2019 = a2015+c2019=a2015+b2017
=> b2017 = a2015 = c2019
=>S=\(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}=\frac{2a^{2015}}{a^{2015}}+\frac{2b^{2017}}{b^{2017}}+\frac{2c^{2019}}{c^{2019}}=2+2+2=6\)
VẬY S ko phụ thuộc vào các giá trị của a,b,c
từ 2 trường hợp trên => giá trị của biểu thức S ko phụ thuộc vào giá trị của a,b,c (đpcm)
\(B=\frac{1}{2017}+\frac{2}{2017}+...+\frac{2019}{2017}\)
\(B=\frac{1+2+...+2019}{2017}\)
Xét tử số có 2019 số hạng
Giá trị ( tổng ) của tử số là : \(\left(1+2019\right)\cdot2019:2=2039190\)
\(\Rightarrow B=\frac{2039190}{2017}=1011,001487\)