( toán 10)- hệ phương trình....

H

hoangtrongminhduc

$Do \ y \ khac \ 0 \ nen \ pt1<=>y\sqrt{x-2y}+2y+6y^2=x<=> x-2y-y\sqrt{x-2y}+\frac{1}{4}y^{2}-\frac{25}{4}y^{2}=0 \\ <=> (\sqrt{x-2y}+2y)(\sqrt{x-2y}-3y)=0 <=> x=4y^2+2y (1)\ V \ x=9y^2+2y(2) \\ thay \ (1) \ va (2) \ vao \ pt2 \ roi \ dat \ an \ phu \ giai$
 
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