HPT: $\left\{\begin{matrix} y^{2}=(x+8)(x^{2}+2) & & \\ 16x-8y+16=5x^{2}+4xy-y^{2}& & \end{matrix}\right.$
$\left\{\begin{matrix} y^{2}=(x+8)(x^{2}+2) & & \\ 16x-8y+16=5x^{2}+4xy-y^{2}& & \end{matrix}\right.$
#1
Posted 17-03-2016 - 22:37
What is .......>_<.....
#3
Posted 17-03-2016 - 23:23
HPT: $\left\{\begin{matrix} y^{2}=(x+8)(x^{2}+2) & & \\ 16x-8y+16=5x^{2}+4xy-y^{2}& & \end{matrix}\right.$
Ta có $PT(2) \Leftrightarrow 5x^2+4xy-y^2-16x+8y-16=0$
$\Leftrightarrow (5x-y+4)(x+y-4)=0$
$\Leftrightarrow \begin{bmatrix}y=5x+4 &\\ y=4-x \end{bmatrix}$
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Case 1:$y=4-x$
Thay vào PT(1) ta có:
$(x-4)^2=(x+8)(x^2+2)$
$\Leftrightarrow... x(x+2)(x+5)=0$
$\Leftrightarrow \begin{bmatrix}x=0,y=4-x=4& & \\ x=-2,y=6 & & \\ x=-5,y=9\end{bmatrix}$
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Case 2:$y=5x+4$
Thay vào PT(1) ta có:
$(5x+4)^2=(x+8)(x^2+2)$
$\Leftrightarrow...x(19-x)(x+2)=0$
$\Leftrightarrow \begin{bmatrix}x=0,y=5x+4=4& & \\ x=-2,y=-6 & & \\ x=19,y=99\end{bmatrix}$
Vậy hệ phương trình có nghiệm $(x,y)=(-5,9);(-2,$$\pm 6$$);(0,4);(19,99)$
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