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Ta có: \(\left(2018+2017\right)^2>2018^2+2017^2\)
Ta có: \(C=\frac{2018^2-2017^2}{2018^2+2017^2}\)
\(=\frac{\left(2018-2017\right)\left(2018+2017\right)}{2018^2+2017^2}=\frac{2018+2017}{2018^2+2017^2}\)
Ta có: \(D=\frac{2018-2017}{2018+2017}\)
\(=\frac{\left(2018-2017\right)\left(2018+2017\right)}{\left(2018+2017\right)^2}=\frac{2018+2017}{\left(2018+2017\right)^2}\)
Đặt a=2018
b=2017
Ta có: \(\left(2018+2017\right)^2=\left(a+b\right)^2\)
\(2018^2+2017^2=a^2+b^2\)
mà \(\left(2018+2017\right)^2>2018^2+2017^2\)(cmt)
nên \(\left(a+b\right)^2>a^2+b^2\)
\(\Leftrightarrow\frac{a+b}{\left(a+b\right)^2}< \frac{a+b}{a^2+b^2}\)
hay \(\frac{2018+2017}{\left(2018+2017\right)^2}< \frac{2018+2017}{2018^2+2017^2}\)
hay D<C
Ta có \(A=\frac{2017-2018}{2017+2018}=\frac{\left(2017-2018\right)\left(2017+2018\right)}{\left(2017+2018\right)^2}=\frac{2017^2-2018^2}{2017^2+2018^2+2.2017.2018}< \frac{2017^2-2018^2}{2017^2+2018^2}=B\)
Vậy A<B
Ta thấy \(A=\frac{2018-2017}{2018+2017}=\frac{2018^2-2017^2}{\left(2018+2017\right)^2}=\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}\)
Mà \(2018^2+2.2018.2017+2017^2>2018^2+2017^2\)
\(\Rightarrow\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}< \frac{2018^2-2017^2}{2018^2+2017^2}\)
Vậy A<B
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2=0-2\cdot0\)
\(\Rightarrow a=b=c=0\)
Thế kết quả vào: \(\left(0-2017\right)^{2018}+\left(0-2017\right)^{2018}-\left(0+2017\right)^{2018}=2017^{2018}\)
Ps: \(\left(-2017\right)^{2018}=2017^{2018}\)
ta có 2015 x 2017 >2017^2 -2
2016 x 2018 > 2016^2
=> A> B
