a) Ta có :
$\begin{array}{l}
2{y^2} + 2{z^2} + 2yz = 2 - 3{x^2}\\
\Leftrightarrow {(x + y + z)^2} + {(x - y)^2} + {(x - z)^2} = 2\\
\Rightarrow {(x + y + z)^2} \le 2 \Rightarrow - \sqrt 2 \le x + y + z \le \sqrt 2
\end{array}$
Vậy Min P = $ - \sqrt 2 $ ; Max P = $\sqrt 2 $
b) Ta có : $1 \ge a + \dfrac{1}{b} \ge 2.\sqrt {\dfrac{a}{b}} \Rightarrow \dfrac{a}{b} \le \dfrac{1}{4} \Rightarrow \dfrac{b}{a} \ge 4$
Lại có:$\dfrac{a}{b} + \dfrac{b}{a} = (\dfrac{a}{b} + \dfrac{b}{{16a}}) + \dfrac{{15b}}{{16a}} \ge 2.\sqrt {\dfrac{{a.b}}{{b.16a}}} + \dfrac{{15.4}}{{16}} = \dfrac{{17}}{4}$
=> Min B = 17/4 khi $16{a^2} = {b^2} = > 4a = b = > {(2a - 1)^2} \le 0 = > a = \dfrac{1}{2};b = 2$
c) $\begin{array}{l}
C = \dfrac{{{{(x + 3y)}^2}}}{{{x^2} + {y^2}}} - 1 \ge - 1\\
C = 9 - \dfrac{{{{(x + 3y)}^2}}}{{{x^2} + {y^2}}} \le 9
\end{array}$
=> Min C = -1 ; Max C = 9