-(-219)+(-219)-401+12

https://olm.vn/hoi-dap/detail/108515110153.html

=> \(\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)

\(\Rightarrow\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)

đoạn tiếp tham khảo tại: Boul đz :D

Ta có \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+3a^2b+3ab^2+c^3-3abc-3a^2b-3ab^2=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(\Rightarrow M=\frac{a^{2019}}{b^{2019}}+\frac{b^{2019}}{c^{2019}}+\frac{c^{2019}}{a^{2019}}=\frac{a^{2019}}{a^{2019}}+\frac{b^{2019}}{b^{2019}}+\frac{c^{2019}}{c^{2019}}=1+1+1=3\)

Đặt \(S = \sqrt{2019 - \sqrt{4037}} - \sqrt{2019 + \sqrt{4037}}\)

=>S<0

\(S = \sqrt{2019 - \sqrt{4037}} - \sqrt{2019 + \sqrt{4037}}\)

=>\(S^2=2019-\sqrt{4037}+2019+\sqrt{4037}-2\cdot\sqrt{\left(2019-\sqrt{4037}\right)\left(2019+\sqrt{4037}\right)}\)

=>\(S^2 = 4038 - 2 \sqrt{2019^2 - 4037}\)

=>\(S^2 = 4038 - 2 \sqrt{2018^2} = 4038 - 2 \cdot 2018 = 4038 - 4036 = 2\)

=>\(S=-\sqrt2\)

=>a+b+c=0

=>a+b=-c

\(A=a^3+a^2c-abc+b^2c+b^3+20182019\)

\(=(a^3+b^3)+(a^2c-abc+b^2c)+20182019\)

\(=(a+b)(a^2-ab+b^2)+c(a^2-ab+b^2)+20182019\)

\(=(a^2-ab+b^2)(a+b+c)+20182019\)

=20182019