làm cái đề ra ấy, ngại viết lại đề :P

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=0\)

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

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

\(\Rightarrow M=1^{2018}+1^{2019}+1^{2020}=1+1+1=3\)

\(a+b+c=9\)

\(\Leftrightarrow\)\(\left(a+b+c\right)^2=81\)

\(\Leftrightarrow\)\(a^2+b^2+c^2+2ab+2bc+2ca=81\)

\(\Leftrightarrow\)\(2\left(ab+bc+ca\right)=54\)

\(\Leftrightarrow\)\(ab+bc+ca=27\)

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

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

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

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

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

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

\(\Rightarrow\)\(B=\left(a-4\right)^{2018}+\left(b-4\right)^{2019}+\left(c-4\right)^{2020}=4^{2018}-4^{2019}+4^{2020}\)

\(\Rightarrow\)\(B=13.4^{2018}\)

Vậy \(B=13.4^{2018}\)

Chúc bạn học tốt ~ 

Phùng Minh Quân : sửa dòng thứ 4 từ dưới lên

Mà \(a+b+c=9\)

\(\Rightarrow a=b=c=3\)

\(B=\left(a-4\right)^{2018}+\left(b-4\right)^{2019}+\left(c-4\right)^{2020}\)

\(B=\left(3-4\right)^{2018}+\left(3-4\right)^{2019}+\left(3-4\right)^{2020}\)

\(B=\left(-1\right)^{2018}+\left(-1\right)^{2019}+\left(-1\right)^{2020}\)

\(B=1-1+1\)

\(B=1\)

Ta có: \(a+b+c=9\)

\(\Leftrightarrow\left(a+b+c\right)^2=81\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=81\)

\(\Leftrightarrow27+2ab+2bc+2ac=81\)

\(\Leftrightarrow2\left(ab+bc+ac\right)=54\)

\(\Leftrightarrow ab+bc+ac=27\)

mà \(a^2+b^2+c^2=27\)

nên \(a^2+b^2+c^2=ab+ac+bc\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)

mà a+b+c=9

nên a=b=c=3

Ta có: \(B=\left(a-4\right)^{2018}+\left(b-4\right)^{2019}+\left(c-4\right)^{2020}\)

\(=\left(3-4\right)^{2018}+\left(3-4\right)^{2019}+\left(3-4\right)^{2020}\)

\(=\left(-1\right)^{2018}-1^{2019}+\left(-1\right)^{2020}\)

\(=1-1+1\)

\(=1\)

Vậy: B=1

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

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

A= 1²⁰¹⁷ + 0²⁰¹⁸ + (-1)²⁰¹⁹ = 0

-Tham khảo:

https://hoc24.vn/cau-hoi/cho-abc-la-cac-so-thoa-man-2018le-abcle2019-tim-gtln-cua-bieu-thuc-plefta-bright2000leftb-cright2000leftc-aright.253535226325

Không mất tính tổng quát giả sử \(a\ge b\ge c\)

đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}0\le x;y;z\le1\\x+y=z\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x^{2000}\le x\\y^{2000}\le y\\z^{2000}\le z\end{matrix}\right.\)

\(\Rightarrow P=x^{2000}+y^{2000}+z^{2000}\le x+y+z=2z\le2\)

\(\Rightarrow P_{max}=1\) khi (x;y;z)=(0;1;1) và hoán vị

\(\Rightarrow\left(a;b;c\right)=\left(2018;2018;2019\right)\) và hoán vị

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

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

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

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

Khi a + b = 0

=> (a + b)(b + c)(c + a) = 0 (2)

Nếu a + b \(\ne0\)

=> ab = -(a + b + c).c

=> ab + (a + b + c).c = 0

=> ab + ac + bc + c² = 0

=> (a + c)(b + c) = 0

=> (a + b)(b + c)(a + c) = 0 (1)

Từ (2)(1) => (a + b)(b + c)(a + c) = 0 \(\forall a;b;c\)

=> a = -b hoặc b = -c hoặc = c = -a

Nếu a = -b => a¹¹ = -b¹¹ => a¹¹ + b¹¹ = 0

=> P = 0 (3)

Nếu b = -c => b⁹ = - c⁹ => b⁹ + c⁹ = 0

=>P = 0 (4)

Nếu c = -a => c²⁰⁰¹ = -a²⁰⁰¹ => c²⁰⁰¹ + a²⁰⁰¹ = 0

=> P = 0 (5)

Từ (3);(4);(5) => P = 0 trong cả 3 trường hợp 

Vạy P = 0

Xyz là ad ak?