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

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

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

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

Do : \(abc=2018\)nên : \(a,b,c\ne0\)

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

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

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

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

\(\frac{1}{2}\)

\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)

\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)

\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)

\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)

\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)

\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)

\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)

\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)

\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)

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

M = \(\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

M = \(\dfrac{a^2bc}{ab\left(ac+c+1\right)}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}\)

M= \(\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

M = \(\dfrac{ac+c+1}{ac+c+1}\)

M = 1

Số 2018 kia như đang gợi ý cho bài này vậy :)

\(M=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}=\dfrac{a^2bc}{ab\left(ac+c+1\right)}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}=\dfrac{ac+c+1}{ac+c+1}=1\)

\(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2=0\)

Mà \(a^2;b^2;c^2\ge0\forall a;b;c\) nên điều này xảy ra \(\Leftrightarrow a=b=c=0\)

\(\Rightarrow M=2018^{2014}+2018^{2014}-2018^{2014}=2018^{2014}\)