( toán 10) - hệ phương trình.

N

noinhobinhyen

pt (1) $x^4+2x^3y+x^2y^2 = 2x+9 \Leftrightarrow (x^2+xy)^2 = 2x+9$

pt (2) $x^2+2xy=6x+6 \Rightarrow xy=-\dfrac{x^2-6x-6}{2}$ thế lên trên ta có:

$(x^2-\dfrac{x^2-6x-6}{2})^2 = 2x+9$

$\Leftrightarrow (x^2+6x+6)^2 = 8x+36$

$\Leftrightarrow (x^2+6x+6)^2 - 36 = 8x$

$\Leftrightarrow (x^2+6x)(x^2+6x+12)=8x$

$\Leftrightarrow x(x+6)(x^2+6x+12) = 8x$

x=0 ko thoả mãn $\Rightarrow (x+6)(x^2+6x+12) = 8$

$\Rightarrow x=-4 \Rightarrow y=\dfrac{17}{4}$
 
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