Lời giải:

Ta có:

$2(ab+bc+ac)=(a+b+c)^2-(a^2+b^2+c^2)=6^2-12=24=2(a^2+b^2+c^2)$

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

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

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

$\Rightarrow a-b=b-c=c-a=0$

$\Rightarrow a=b=c$. Mà $a+b+c=6$ nên $a=b=c=2$

Khi đó:

$A=(2-3)^{2020}+(2-3)^{2020}+(2-3)^{2020}=1+1+1=3$

\(\hept{\begin{cases}a+b+c=6\left(1\right)\\a^2+b^2+c^2=12\left(2\right)\end{cases}}\)

(1) bình phuong trừ (2)=>ab+bc+ac=12

\(a^2+b^2+c^2\ge ab+bc+ac\)đẳng thức chỉ xẩy ra khi a=b=c

Từ (1)=> a=b=c=2

=> P=3 

a) Áp dụng Cauchy Schwars ta có:

\(M=\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\ge\frac{\left(a+b+c\right)^2}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

b) \(N=\frac{1}{a}+\frac{4}{b+1}+\frac{9}{c+2}\ge\frac{\left(1+2+3\right)^2}{a+b+c+3}=\frac{36}{6}=6\)

Dấu "=" xảy ra khi: x=y=1

c) \(P=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{9}{2.3}=\frac{3}{2}\)

Dấu "=" xảy ra khi: x=y=1

\(\left(a+b+c\right)^2=3a^2+3b^2+3c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

\(\Rightarrow P=a^2+\left(a+2\right)\left(a+a\right)+2020\)

\(\Rightarrow P=3a^2+4a+2020=3\left(a+\frac{2}{3}\right)^2+\frac{6056}{3}\ge\frac{6056}{3}\)

\(P_{min}=\frac{6056}{3}\) khi \(a=-\frac{2}{3}\)

\(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)_{ }\)

\(a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)

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

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

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

\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)

Do đó  \(P=a^2+\left(a+2\right)\left(2a\right)+2020\)

\(P=a^2+2a^2+4a+2020\)

\(P=3a^2+4a+2020\)

\(3P=9a^2+12a+6060\)

\(3P=\left(3a\right)^2+2.\left(3a\right).2+4+6060-4\)

\(3P=\left(3a+2\right)^2+6056\ge6056\Leftrightarrow3P\ge6056\Leftrightarrow P\ge\frac{6056}{3}\)    Dấu "=" xảy ra khi a = b = c = \(-\frac{3}{2}\)

Vậy P đạt giá trị nhỏ nhất là 6056/3 khi a = b = c = -3/2

\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)

\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)

\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)

Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A

Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)

Ta có : a³ + b³ + c³ = 3abc

=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

=> [(a + b)³ + c³] - [(3ab(a + b) + 3abc] = 0

=> (a + b + c)(a² + b² + 2ab - ac - bc + c²) - 3ab(a + b + c) = 0

=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

=> a² + b² + c² - ab- ac - bc = 0

=> 2(a² + b² + c² - ab- ac - bc) = 0

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (b² - 2bc + c²) + (a² - 2ac + c²) = 0

=> (a - b)² + (b - c)² + (a - c)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)

Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

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

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

mà \(a+b+c\ne0\)

nên \(a^2+b^2+c^2-ab-ac-bc=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

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

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

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

Ta có: \(M=\dfrac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}\)

\(=\dfrac{a^{2020}+a^{2020}+a^{2020}}{\left(a+a+a\right)^{2020}}=\dfrac{3\cdot a^{2020}}{9\cdot a^{2020}}=\dfrac{1}{3}\)

Nhầm là, tính A=(a-1)²⁰¹⁹+(b²-1)²⁰²⁰+(c³-1)²⁰²¹

Ta có : \(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)=9-2\times6=3\)

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

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

\(\Rightarrow a=b=c\)

Mà \(a+b+c=3\Rightarrow a=b=c=1\)

\(\Rightarrow A=\left(1-1\right)^{2019}+\left(1^2-1\right)^{2020}+\left(1^3-1\right)^{2021}\)

\(=0^{2019}+0^{2020}+0^{2021}=0\)