Đáp án:
$P = 2$
$A = \dfrac13$
Giải thích các bước giải:
$\begin{array}{l}a) \quad \text{Ta có:}\\ \quad \begin{cases}\dfrac1a + \dfrac1b + \dfrac1c = 2\\a + b+ c = abc \end{cases}\\ \to \begin{cases}\left(\dfrac1a + \dfrac1b + \dfrac1c\right)^2= 4\\\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} = 1 \end{cases}\\ \to \begin{cases}\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + 2\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} \right)= 4\\\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} = 1 \end{cases}\\ \to \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = 4 - 2\cdot 1\\ \to P = 2\\ b)\qquad \text{Ta có:}\\ \quad a^3 +b^3 + c^3 = 3abc\\ \to a^3 + b^3 + c^3 - 3abc = 0\\ \to (a+b+c)(a^2 +b^2 + c^2 - ab - bc - ca) =0\\ \to a^2 +b^2 + c^2 -ab - bc - ca =0\quad (Do\,\,a+b+c\ne0)\\ \to (a-b)^2 + (b-c)^2 + (c-a)^2 = 0\\ \to a =b = c\\ \text{Ta được:}\\ A = \dfrac{a^2 + b^2 + c^2}{(a+b+c)^2}\\ \to A = \dfrac{3a^2}{(3a)^2}\\ \to A = \dfrac{1}{3} \end{array}$
