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datmoc33
2 năm trước
$x²+26y²−10xy+14x−76y+58=0 ⇔x²−10xy+25y²+14x−70y+49+y²−6y+9=0 ⇔(x−5y)²+14(x−5y)+49+(y−3)²=0 ⇔(x−5y+7)²+(y−3)²=0 ⇔x−5y+7=y−3=0 ⇔y=3;x=8$ bài này ngoài cách giải trên còn cách khác k ạ $tìm cặp số x, y thoã mãn x^2+ 26y^2−10xy+14x−76y+58=0$
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