Hệ phương trình

N

nerversaynever

Từ hpt ban đầu suy ra x>1;y>0, biến đổi hpt thành
[TEX]\begin{array}{l}\left\{ \begin{array}{l}x^2 y\left( {\sqrt {y^2 + 1} - \frac{{\sqrt 2 }}{y}} \right) + x^2 \left( {y - 1} \right) = 2\sqrt {x^2 + 4} - \sqrt 2 x^2 + 2x - x^2 \\ 2\left( {\sqrt {y^2 + 3} - 2} \right) = 4x - 4 - \sqrt {4 + 3x^2 } \\ \end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}\frac{{x^2 \left( {y + 1} \right)\left( {y^2 + 1} \right)\left( {y - 1} \right)}}{{y\sqrt {y^2 + 1} + \sqrt 2 }} + x^2 \left( {y - 1} \right) = - 2\frac{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x^2 + 2} \right)}}{{2\sqrt {x^2 + 4} + \sqrt 2 x^2 }} - x\left( {x - 2} \right) \\ 2\frac{{y^2 - 1}}{{\sqrt {y^2 + 3} + 2}} = \frac{{\left( {x - 2} \right)\left( {13x - 6} \right)}}{{4x - 4 + \sqrt {4 + 3x^2 } }} \\ \end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l}A\left( {y - 1} \right) = - B\left( {x - 2} \right) \\ C\left( {y - 1} \right) = D\left( {x - 2} \right) \\ \end{array} \right. \Rightarrow \left( {AD + BC} \right)\left( {y - 1} \right)\left( {x - 2} \right) = 0 \\ \end{array}[/TEX]
vơi A,B,C,D>0
suy ra hpt có no x=2;y=1
 
U

umbala1996

Mình cũng ra vậy. Cảm ơn bạn nhé. :)
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