Đáp án:
a. \(\left\{ \begin{array}{l}
x = \dfrac{7}{{4 + \sqrt 3 }}\\
y = \dfrac{{10 - \sqrt 3 }}{{4 + \sqrt 3 }}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x\sqrt 3 + my = 5\\
mx - y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 1\\
x\sqrt 3 + {m^2}x - m = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 1\\
\left( {\sqrt 3 + {m^2}} \right)x = 5 + m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 5}}{{\sqrt 3 + {m^2}}}\\
y = \dfrac{{{m^2} + 5m - \sqrt 3 - {m^2}}}{{\sqrt 3 + {m^2}}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 5}}{{\sqrt 3 + {m^2}}}\\
y = \dfrac{{5m - \sqrt 3 }}{{\sqrt 3 + {m^2}}}
\end{array} \right.\\
a.Thay:m = 2\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2 + 5}}{{\sqrt 3 + {2^2}}}\\
y = \dfrac{{5.2 - \sqrt 3 }}{{\sqrt 3 + {2^2}}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{7}{{4 + \sqrt 3 }}\\
y = \dfrac{{10 - \sqrt 3 }}{{4 + \sqrt 3 }}
\end{array} \right.\\
b.Do:{m^2} \ge 0\left( {ld} \right)\forall x \in R\\
\to \sqrt 3 + {m^2} > 0\left( {ld} \right)\forall x \in R\\
\to dpcm
\end{array}\)
