<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)

a=b=c => 3²⁰²⁰ = 3.a²⁰¹⁹ <=> 3²⁰¹⁹ = a²⁰¹⁹ => a=b=c=3

A= 1²⁰¹⁷ + 0²⁰¹⁸ + (-1)²⁰¹⁹ = 0

\(\left(a+b+c\right)\left(ab+bc+ca\right)=abc\)

\(\Rightarrow\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc\right)-abc=0\)

\(\Rightarrow a^2b+bc^2+2abc+a^2c+ac^2+b^2c+ab^2=0\)

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

\(\Rightarrow\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\)

\(\Rightarrow\left(a+c\right)\left(a+b\right)\left(b+c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+c=0\Rightarrow a^{2019}+c^{2019}=0\\b+c=0\Rightarrow b^{2019}+c^{2019}=0\\a+b=0\Rightarrow a^{2019}+b^{2019}=0\end{matrix}\right.\)

\(\Rightarrow P=1\)

*Hằng đẳng thức cần áp dụng:

\(x^n+y^n=\left(x+y\right)\left(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1}\right)\)

nên \(x+y=0\Rightarrow x^n+y^n=0\)

\(a+b=c+\frac{1}{2019}\Leftrightarrow a+b-c=\frac{1}{2019}\Leftrightarrow\frac{1}{a+b-c}=2019\)

\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+2019\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=2019\)

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

\(\Leftrightarrow\frac{a+b}{ab}=\frac{a+b}{c\left(a+b-c\right)}\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)=\left(a+b\right)ab\)

\(\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)-ab\left(a+b\right)=0\)

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

\(\Leftrightarrow\left(a+b\right)\left[c\left(a-c\right)-b\left(a-c\right)\right]=0\)

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

=>a=-b hoặc c=b hoặc a=c

không mất tính tổng quát, giả sử a=-b, ta có:

\(P=\left(-b^{2019}+b^{2019}-c^{2019}\right)\left(-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}-\frac{1}{c^{2019}}\right)=\left(-c\right)^{2019}\cdot\left(\frac{-1}{c}\right)^{2019}=1\)

tương tư với các trường hợp khác ta cũng có P=1

Vậy P=1

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\)

hay \(a^2+b^2+c^2=0\Rightarrow a=b=c=0\)

Thay a = b = c = 0 vào M rồi tính như bình thường nha bạn!

Ta có : 

\(a+b+c=0\)

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

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

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

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

\(\Leftrightarrow\)\(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}\Leftrightarrow a=b=c=0}\)

\(\Rightarrow\)\(M=\left(a-2018\right)^{2019}+\left(b-2018\right)^{2019}-\left(c+2018\right)^{2019}\)

\(\Rightarrow\)\(M=-2018^{2019}-2018^{2019}-2018^{2019}\)

\(\Rightarrow\)\(M=-3.2018^{2019}\)

Chúc bạn học tốt ~