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H

hiensau99

$4x^2+y^2-2x-2xy-y+1=0$

$\to (x^2+y^2-2xy)-2.\dfrac{1}{2}.(y-x)+\dfrac{1}{4}-\dfrac{1}{4}+3x^2-3x+\dfrac{3}{4}-\dfrac{3}{4}+1=0$

$\to (y-x)^2-2.\dfrac{1}{2}.(y-x)+\dfrac{1}{4}+3(x^2-x+\dfrac{1}{4})=0$

$\to (y-x-\dfrac{1}{2})^2+3.(x-\dfrac{1}{2})^2=0$

Tới đây dễ dàng tìm $x=\dfrac{1}{2}; y=1$
 
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