a) Ta có:
\(\left\{ \matrix{ x\sqrt 2 - y\sqrt 3 = 1 \hfill \cr x + y\sqrt 3 = \sqrt 2 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x\sqrt 2 - y\sqrt 3 = 1 \hfill \cr x = \sqrt 2 - y\sqrt 3 \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
\left( {\sqrt 2-y\sqrt 3 } \right)\sqrt 2 - y\sqrt 3 = 1 \ (1) \hfill \cr
x = \sqrt 2 - y\sqrt 3 \ (2) \hfill \cr} \right.\)
Giải phương trình \((1)\), ta được:
\(( \sqrt 2 - y\sqrt 3)\sqrt 2 - y\sqrt 3 = 1\)
\( \Leftrightarrow (\sqrt 2)^2 - y\sqrt 3 . \sqrt 2 - y\sqrt 3 = 1 \)
\( \Leftrightarrow 2 - y\sqrt 3 . \sqrt 2 - y\sqrt 3 = 1 \)
\( \Leftrightarrow -y\sqrt 3. \sqrt 2 - y\sqrt 3 = 1 - 2\)
\( \Leftrightarrow -y\sqrt 3 (\sqrt 2 +1) = -1 \)
\( \Leftrightarrow y=\dfrac{1}{\sqrt 3(\sqrt 2 +1)}=\dfrac{\sqrt 3 (\sqrt 2 -1)}{(\sqrt 3)^2(\sqrt 2 +1)(\sqrt 2 -1)}\)
\( \Leftrightarrow y= \dfrac{\sqrt 3 (\sqrt 2 -1)}{3}\)
Thay \(y\) tìm được vào phương trình \((2)\), ta được:
\(x = \sqrt 2 - \dfrac{\sqrt 3 (\sqrt 2 -1)}{3}.\sqrt 3\)
\( \Leftrightarrow x=\sqrt 2 - \dfrac{\sqrt 3 .\sqrt 3(\sqrt 2 -1)}{3} \)
\(\Leftrightarrow x=\sqrt 2 - \dfrac{ 3(\sqrt 2 -1)}{3} =\sqrt 2 - (\sqrt 2 -1) \)
\(\Leftrightarrow x=\sqrt 2 -\sqrt 2 +1=1.\)
Vậy hệ phương trình đã cho có nghiệm duy nhất là: \( {\left( 1;\dfrac{\sqrt 3 (\sqrt 2 -1)}{3} \right)}\)
b) Ta có:
\(\left\{ \matrix{ x - 2\sqrt 2 y = \sqrt 5 \hfill \cr x\sqrt 2 + y = 1 - \sqrt {10} \hfill \cr} \right.\)
\(\Leftrightarrow \left\{ \matrix{
x = 2\sqrt 2 y + \sqrt 5 \ (1) \hfill \cr
\left( {2\sqrt 2 y + \sqrt 5 } \right).\sqrt 2 + y = 1 - \sqrt {10}\ (2) \hfill \cr} \right.\)
Giải phương trình \((2)\), ta được:
\(\left( {2\sqrt 2 y + \sqrt 5 } \right).\sqrt 2 + y = 1 - \sqrt {10}\)
\(\Leftrightarrow 2(\sqrt 2 .\sqrt 2)y + \sqrt 5 .\sqrt 2 + y = 1 - \sqrt {10}\)
\(\Leftrightarrow 4y + \sqrt{10}+y=1- \sqrt{10}\)
\(\Leftrightarrow 4y +y=1- \sqrt{10}- \sqrt{10} \)
\(\Leftrightarrow 5y=1-2 \sqrt{10}\)
\(\Leftrightarrow y=\dfrac{1-2 \sqrt{10}}{5}\)
Thay \(y=\dfrac{1-2 \sqrt{10}}{5}\) vào \((1)\), ta được:
\(x = 2\sqrt 2 .\dfrac{1-2 \sqrt{10}}{5} + \sqrt 5= \dfrac{2\sqrt 2 -4 \sqrt{20}}{5} + \sqrt 5\)
\(\Leftrightarrow x=\dfrac{2\sqrt 2 -4 .2\sqrt{5}}{5} + \sqrt 5=\dfrac{2\sqrt 2 -8\sqrt{5}+ 5\sqrt 5}{5}\)
\(\Leftrightarrow x=\dfrac{2 \sqrt 2 -3 \sqrt 5}{5}\)
Vậy hệ có nghiệm duy nhất là: \((x; y)\) = \({\left(\dfrac{2\sqrt{2} - 3\sqrt{5}}{5};\dfrac{1 - 2\sqrt{10}}{5}\right)}\);
c) Ta có:
\(\left\{ \matrix{ \left( {\sqrt 2 - 1} \right)x - y = \sqrt 2 \hfill \cr x + \left( {\sqrt 2 + 1} \right)y = 1 \hfill \cr} \right. \)
\(\left\{ \begin{array}{l}y = \left( {\sqrt 2 - 1} \right)x - \sqrt 2 \,\,\,\,\,\left( 1 \right)\\x + \left( {\sqrt 2 + 1} \right)\left[ {\left( {\sqrt 2 - 1} \right)x - \sqrt 2 } \right] = 1\,\,\,\left( 2 \right)\end{array} \right.\)
Giải phương trình \((2)\), ta được:
\(x + \left( {\sqrt 2 + 1} \right)\left[ { \left( {\sqrt 2 - 1} \right)x} -\sqrt 2 \right] = 1\)
\(\Leftrightarrow x + (\sqrt 2 + 1) (\sqrt 2 - 1)x -( \sqrt 2 + 1). \sqrt 2 = 1\)
\(\Leftrightarrow x + {\left((\sqrt 2)^2 - 1^2 \right)}x-( 2 + \sqrt 2) = 1\)
\(\Leftrightarrow x + x = 1+( 2 + \sqrt 2)\)
\(\Leftrightarrow 2x =3 +\sqrt 2\)
\(\Leftrightarrow x=\dfrac{3+ \sqrt 2}{2}\)
Thay \(x=\dfrac{3+ \sqrt 2}{2}\) vào \((1)\), ta được:
\(y = \left( {\sqrt 2 - 1} \right).\dfrac{3+ \sqrt 2}{2} - \sqrt 2\)
\( \Leftrightarrow y= \dfrac{(\sqrt 2 - 1 )(3+ \sqrt 2)}{2} - \sqrt 2 \)
\( \Leftrightarrow y= \dfrac{3\sqrt 2 -3 +2 -\sqrt 2}{2} - \sqrt 2 \)
\( \Leftrightarrow y= \dfrac{2\sqrt 2 -1}{2} - \sqrt 2 \)
\( \Leftrightarrow y= \dfrac{2\sqrt 2 -1-2\sqrt 2}{2} \)
\( \Leftrightarrow y= \dfrac{-1}{2} \)
Vậy hệ có nghiệm \((x; y) = {\left(\dfrac{3 + \sqrt{2}}{2};\dfrac{-1}{2} \right)}\)
