Đáp án:
B
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
y = 2x - k - 1\\
x + 2x - k - 1 = 2k - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 2x - k - 1\\
3x = 3k
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = k\\
y = 2k - k - 1 = k - 1
\end{array} \right.\\
P = {x^2} + {y^2} = {k^2} + {\left( {k - 1} \right)^2}\\
= {k^2} + {k^2} - 2k + 1\\
= 2{k^2} - 2k + 1\\
= 2{k^2} - 2.k\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{1}{2}\\
= {\left( {k\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2}\\
Do:{\left( {k\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall k\\
\to {\left( {k\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\to Min = \dfrac{1}{2}\\
\Leftrightarrow k\sqrt 2 - \dfrac{1}{{\sqrt 2 }} = 0\\
\Leftrightarrow k = \dfrac{1}{2}
\end{array}\)
