Đáp án:
$\begin{array}{l}
a)Khi:y = 3\\
\Rightarrow {x^2} = 3\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.\\
\Rightarrow \left( {\sqrt 3 ;3} \right);\left( { - \sqrt 3 ;3} \right) \in \left( P \right)\\
b)O\left( {0;0} \right) \in \left( d \right)\\
\Rightarrow 0 = k.0 - k + 1\\
\Rightarrow - k + 1 = 0\\
\Rightarrow k = 1\\
c)Xet:{x^2} = kx - k + 1\\
\Rightarrow {x^2} - kx + k - 1 = 0\\
\Rightarrow \Delta = {k^2} - 4k + 4 = {\left( {k - 2} \right)^2} \ge 0
\end{array}$
=> chúng luôn cắt nhau tại 2 điểm phân biêt
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = k\\
{x_1}{x_2} = k - 1
\end{array} \right.\\
M = \dfrac{{2{x_1}{x_2} + 3}}{{x_1^2 + x_2^2 + 2\left( {{x_1}{x_2} + 1} \right)}}\\
= \dfrac{{2\left( {k - 1} \right) + 3}}{{{{\left( {{x_1} + {x_2}} \right)}^2} + 2}}\\
= \dfrac{{2k + 1}}{{{k^2} + 2}}\\
\Rightarrow M.{k^2} + 2M = 2k + 1\\
\Rightarrow M.{k^2} - 2k + 2M - 1 = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow 1 - M.\left( {2M - 1} \right) \ge 0\\
\Rightarrow 2{M^2} - M - 1 \le 0\\
\Rightarrow \left( {2M + 1} \right)\left( {M - 1} \right) \le 0\\
\Rightarrow - \dfrac{1}{2} \le M \le 1\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:M = - \dfrac{1}{2}\,khi:k = - 2\\
GTLN:M = 1\,khi:k = 1
\end{array} \right.
\end{array}$
