mk nhầm đề bài là: a^2017+b^2017=2a^2018.b^2018

giải s z

\(ac=b^2\)

=>\(\frac{a}{b}=\frac{b}{c}\)

\(ab=c^2\)

=>\(\frac{b}{c}=\frac{c}{a}\)

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

=>a=b=c

\(M=\frac{a^{2017}}{b^{2018}\cdot c^{2019}}=\frac{a^{2017}}{a^{2018}\cdot a^{2019}}=\frac{a^{2017}}{a^{4037}}=\frac{1}{a^{2020}}\)

\(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)

\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{a\left(bc+b+2018\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)

\(\Rightarrow M=\frac{2018a}{ab+2018a+2018}+\frac{ab}{ab+2018a+2018}+\frac{1}{ab+2018a+2018}\)

\(\Rightarrow M=\frac{2018a+ab+1}{2018a+ab+1}=1\)

Do : \(abc=2018\)nên : \(a,b,c\ne0\)

Ta có : \(M=\frac{2018a}{ab+2018a+2018}+\frac{b}{bc+b+2018}+\frac{c}{ac+c+1}\)

\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{abc+ab+2018a}+\frac{abc}{a^2bc+abc+ab}\)

\(=\frac{2018a}{ab+2018a+2018}+\frac{ab}{2018+ab+2018a}+\frac{2018}{2018+ab+2018a}\)

\(=\frac{2018a+ab+2018}{ab+2018a+2018}=1\)

Chú ý: \(\frac{ab}{a+b}\le\frac{\frac{\left(a+b\right)^2}{4}}{a+b}=\frac{a+b}{4}\).

Tương tự và cộng theo vế thu được \(M\le\frac{a+b+c}{2}=\frac{2018}{2}=1009\)

Đẳng thức xảy ra khi a = b = c = ²⁰¹⁸/₃