Предмет: Математика, автор: fhdb2Like

Решите систему. Номер 437

Ответы

Автор ответа: HIPM

Решение:
$left { {{(x-2)(y+3)=160} atop {y-x=1}} right.\ left { {{(x-2)(y+3)=160} atop {y=1+x}} right. \(x-2)(1+x+3)=160\4x+ x^{2} -8-2x=160\2x+-8-160=0\ x^{2} +2x-168=0\D1=1+168=169\x1=12\x2=-14\y1=13\y2=-13\Answer: (12;13), (-14;-13)\\ left { {{(x-1)(y+10)=9} atop {x-y=11}} right. \ left { {{(x-1)(y+10)=9} atop {x=11+y}} right. \(11+y-1)(y+10)=9\(10+y)(10+y)=9\10+y=3\y1=-7\y2=-13\x1=4\x2=-2\Answer: (4;-7), (-2;-13)$
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Answer = Ответ

Автор ответа: fhdb2Like

Спасибо огромное, всех благ тебе и богатства

Автор ответа: Prescure

незачто, обращайтесь, любезнейший!

Автор ответа: HIPM

Ах ты)

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