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Giải các hệ phương trình sau:
\(a)\left\{ {\matrix{
{\sqrt 3 x – 2\sqrt 2 y = 7} \cr
{\sqrt 2 x + 3\sqrt 3 y = – 2\sqrt 6 } \cr} } \right.\)
\(b)\left\{ {\matrix{
{\left( {\sqrt 2 + 1} \right)x – \left( {2 – \sqrt 3 } \right)y = 2} \cr
{\left( {2 + \sqrt 3 } \right)x + \left( {\sqrt 2 – 1} \right)y = 2} \cr} } \right.\)
a)
\(\eqalign{
& \left\{ {\matrix{
{\sqrt 3 x – 2\sqrt 2 y = 7} \cr
{\sqrt 2 x + 3\sqrt 3 y = – 2\sqrt 6 } \cr
} } \right. \Leftrightarrow \left\{ {\matrix{
{\sqrt 6 x – 4y = 7\sqrt 2 } \cr
{\sqrt 6 x + 9y = – 6\sqrt 2 } \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{13y = – 13\sqrt 2 } \cr
{\sqrt 3 x – 2\sqrt 2 y = 7} \cr
} } \right. \Leftrightarrow \left\{ {\matrix{
{y = – \sqrt 2 } \cr
{\sqrt 3 x – 2\sqrt 2 .\left( { – \sqrt 2 } \right) = 7} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{y = – \sqrt 2 } \cr
{\sqrt 3 x = 3} \cr
} } \right. \Leftrightarrow \left\{ {\matrix{
{y = – \sqrt 2 } \cr
{x = \sqrt 3 } \cr} } \right. \cr} \)
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Vậy hệ phương trình đã cho có một nghiệm (x; y) = \(\left( {\sqrt 3 ; – \sqrt 2 } \right)\)
b)
\(\eqalign{
& \left\{ {\matrix{
{\left( {\sqrt 2 + 1} \right)x – \left( {2 – \sqrt 3 } \right)y = 2} \cr
{\left( {2 + \sqrt 3 } \right)x + \left( {\sqrt 2 – 1} \right)y = 2} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 – 1} \right)x – \left( {\sqrt 2 – 1} \right)\left( {2 – \sqrt 3 } \right)y = 2\left( {\sqrt 2 – 1} \right)} \cr
{\left( {2 + \sqrt 3 } \right)\left( {2 – \sqrt 3 } \right)x + \left( {2 – \sqrt 3 } \right)\left( {\sqrt 2 – 1} \right)y = 2\left( {2 – \sqrt 3 } \right)} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{x – \left( {\sqrt 2 – 1} \right)\left( {2 – \sqrt 3 } \right)y = 2\left( {\sqrt 2 – 1} \right)} \cr
{x + \left( {2 – \sqrt 3 } \right)\left( {\sqrt 2 – 1} \right)y = 2\left( {2 – \sqrt 3 } \right)} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{2x = 2\sqrt 2 – 2 + 4 – 2\sqrt 3 } \cr
{x + \left( {2 – \sqrt 3 } \right)\left( {\sqrt 2 – 1} \right)y = 2\left( {2 – \sqrt 3 } \right)} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{x = \sqrt 2 + 1 – \sqrt 3 } \cr
{\left( {2 – \sqrt 3 } \right)\left( {\sqrt 2 – 1} \right)y = 4 – 2\sqrt 3 – \sqrt 2 – 1 + \sqrt 3 } \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{x = \sqrt 2 + 1 – \sqrt 3 } \cr
{y = {{3 – \sqrt 2 – \sqrt 3 } \over {\left( {2 – \sqrt 3 } \right)\left( {\sqrt 2 – 1} \right)}}} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{x = \sqrt 2 + 1 – \sqrt 3 } \cr
{y = {{\left( {3 – \sqrt 2 – \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)\left( {\sqrt 2 + 1} \right)} \over {\left( {2 – \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)\left( {\sqrt 2 – 1} \right)\left( {\sqrt 2 + 1} \right)}}} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{x = \sqrt 2 + 1 – \sqrt 3 } \cr
{y = {{\left( {3 – \sqrt 2 – \sqrt 3 } \right)\left( {2\sqrt 2 + \sqrt 6 + 2 + \sqrt 3 } \right)} \over {\left( {4 – 3} \right)\left( {2 – 1} \right)}}} \cr
} } \right. \cr
& \Leftrightarrow \left\{ {\matrix{
{x = \sqrt 2 + 1 – \sqrt 3 } \cr
{y = \sqrt 2 – 1 – \sqrt 3 } \cr} } \right. \cr} \)
Vậy hệ phương trình đã cho có một nghiệm (x; y) = \(\left( {\sqrt 2 + 1 – \sqrt 3 ;\sqrt 2 – 1 – \sqrt 3 } \right)\)