Câu 9:
\({m^2}\left( {x + m} \right) = x + m \Leftrightarrow \left( {{m^2} - 1} \right)x + {m^3} - m = 0\)
Phương trình có tập nghiệm \(\mathbb{R}\) \( \Leftrightarrow \left\{ \begin{array}{l}{m^2} - 1 = 0\\{m^3} - m = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = \pm 1\\m = 0,m = \pm 1\end{array} \right. \Leftrightarrow m = \pm 1\)
Câu 10:
\(\begin{array}{l}\overrightarrow {MD} = \overrightarrow {AD} - \overrightarrow {AM} = \overrightarrow {AD} - \dfrac{1}{{\sqrt 2 }}\overrightarrow {AB} \\\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {AD} \\ \Rightarrow \overrightarrow {MD} .\overrightarrow {AC} = \left( {\overrightarrow {AD} - \dfrac{1}{{\sqrt 2 }}\overrightarrow {AB} } \right)\left( {\overrightarrow {AB} + \overrightarrow {AD} } \right)\\ = \overrightarrow {AD} .\overrightarrow {AB} - \dfrac{1}{{\sqrt 2 }}A{B^2} + A{D^2} - \dfrac{1}{{\sqrt 2 }}\overrightarrow {AB} .\overrightarrow {AD} \\ = 0 - \dfrac{1}{{\sqrt 2 }}.2{a^2} + {a^2} - \dfrac{1}{{\sqrt 2 }}.0\\ = \left( {1 - \sqrt 2 } \right){a^2}\end{array}\)
