Giải hệ :
$\left\{\begin{matrix} 2y^2-x^2=1&&\\2x^3-y^3=2y-x&&\end{matrix}\right.$
$\left\{\begin{matrix} 2y^2-x^2=1&&\\2x^3-y^3=2y-x&&\end{matrix}\right.$
Started By dorabesu, 26-01-2013 - 19:53
#1
Posted 26-01-2013 - 19:53
dorabesu
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Trung sĩ
#2
Posted 26-01-2013 - 20:07
vutuanhien
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- ĐHV Toán Cao cấp
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Thiếu úy
Từ phương trình thứ nhất suy ra $2x^3-y^3=(2y-x)(2y^2-x^2)\Rightarrow x^3-5y^3+2x^2y+2xy^2=0\Rightarrow (x-y)\left [x^2+xy+y^2+4y(x+y)-2xy \right ]=0\Rightarrow (x-y)(x^2+3xy+5y^2)=0\Rightarrow x=y$ hoặc $x^2+3xy+5y^2$=0. Nếu $x^2+3xy+5y^2$=0 thì suy ra x=y=0, thay vào hệ thấy không thỏa mãn. Nếu x=y$\neq$0, thì thay vào hệ pt ta suy ra x=y=$\pm$1Giải hệ :
$\left\{\begin{matrix} 2y^2-x^2=1&&\\2x^3-y^3=2y-x&&\end{matrix}\right.$
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"Algebra is the offer made by the devil to the mathematician. The devil says: I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvelous machine." (M. Atiyah)
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