a) Ta có:
\(\left\{ \matrix{ x + y\sqrt 5 = 0 \hfill \cr x\sqrt 5 + 3y = 1 - \sqrt 5 \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{ x = - y\sqrt 5 \hfill \cr \left( { - y\sqrt 5 } \right).\sqrt 5 + 3y = 1 - \sqrt 5 \hfill \cr} \right.\)
\( \Leftrightarrow \left\{ \matrix{ x = - y\sqrt 5 \hfill \cr - 5y + 3y = 1 - \sqrt 5 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = - y\sqrt 5 \hfill \cr - 2y = 1 - \sqrt 5 \hfill \cr} \right.\)
\( \Leftrightarrow \left\{ \matrix{ x = - y\sqrt 5 \hfill \cr y = \dfrac{1 - \sqrt 5 }{ - 2} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = - y\sqrt 5 \hfill \cr y = \dfrac{\sqrt 5 - 1}{2} \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{ x = - \dfrac{\sqrt 5 - 1}{ 2}.\sqrt 5 \hfill \cr y = \dfrac{\sqrt 5 - 1}{2} \hfill \cr} \right.\)
\( \Leftrightarrow \left\{ \matrix{ x = - \dfrac{5 - \sqrt 5 }{2} \hfill \cr y = \dfrac{\sqrt 5 - 1}{2} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = \dfrac{\sqrt 5 - 5}{ 2} \hfill \cr y = \dfrac{\sqrt 5 - 1}{ 2} \hfill \cr} \right.\)
Vậy hệ phương trình có nghiệm duy nhất \( {\left(\dfrac{\sqrt 5 - 5}{ 2} ; \dfrac{\sqrt 5 - 1}{ 2} \right)}\)
b) Ta có:
\(\left\{ \matrix{ \left( {2 - \sqrt 3 } \right)x - 3y = 2 + 5\sqrt 3 \hfill \cr 4x + y = 4 - 2\sqrt 3 \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
\left( {2 - \sqrt 3 } \right)x - 3\left( {4 - 2\sqrt 3 - 4x} \right) = 2 + 5\sqrt 3 \ (1) \hfill \cr
y = 4 - 2\sqrt 3 - 4x \ (2) \hfill \cr} \right.\)
Giải phương trình \((1)\), ta được:
\(( 2 - \sqrt 3 )x - 3(4 - 2\sqrt 3 - 4x) = 2 + 5\sqrt 3\)
\(\Leftrightarrow 2x -\sqrt 3 x -12 + 6 \sqrt 3 + 12x=2+ 5 \sqrt 3\)
\(\Leftrightarrow 2x -\sqrt 3 x + 12x=2+ 5 \sqrt 3 +12 -6 \sqrt 3 \)
\(\Leftrightarrow (2 -\sqrt 3 + 12)x= 2+12 +5\sqrt 3 -6 \sqrt 3 \)
\(\Leftrightarrow (14- \sqrt 3)x=14-\sqrt 3\)
\(\Leftrightarrow x=1\)
Thay \(x=1\), vào \((2)\), ta được:
\(y = 4 - 2\sqrt 3 - 4.1=-2 \sqrt 3.\)
Vậy hệ phương trình có nghiệm duy nhất \((1; -2 \sqrt 3).\)
