1234

Đáp án:

$B=1.$

Giải thích các bước giải:

$a+b+c=12\\ \Rightarrow (a+b+c)^2=144\\ \Leftrightarrow a^2+b^2+c^2+2(ab+ac+bc)=144\\ \Leftrightarrow 2(ab+ac+bc)=144-(a^2+b^2+c^2)\\ \Leftrightarrow 2(ab+ac+bc)=144-48\\ \Leftrightarrow 2(ab+ac+bc)=96\\ a^2+b^2+c^2=48\\ \Leftrightarrow 2a^2+2b^2+2c^2=96\\ \Leftrightarrow 2a^2+2b^2+2c^2-2(ab+ac+bc)=0\\ \Leftrightarrow 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\ \Leftrightarrow (a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0\\ \Leftrightarrow (a-b)^2+(a-c)^2+(b-c)^2=0$

Mà $(a-b)^2+(a-c)^2+(b-c)^2 \ge 0 \ \forall \ a,b,c$

$\Rightarrow \left\{\begin{array}{l} a-b=0 \\ a-c=0 \\ b-c=0 \end{array} \right.\Leftrightarrow a=b=c$

Mà $a+b+c=12 \Rightarrow a=b=c=4$

$B=(4-3)^{2020}+(3-4)^{2021}+(4-5)^{2022}\\ =1^{2020}+(-1)^{2021}+(-1)^{2022}\\ =1-1+1\\ =1.$