Решаем при помощи замены: пусть у=x²+4x, тогда у·(у-17)+60=0;
у²-17у+60=0. Решаем уравнени по т. Виета; у₁=12, у₂=5.
Возвращаемся к замене:
1)x²+4x=12;
x²+4x-12=0, решаем по т. Виета: х₁=2, х₂=-6.
2)x²+4x=5;
x²+4x-5;решаем по т. Виета: х₁=1, х₂=-5.
Ответ:1; 2;-5;-6.
Автор:
boozerwoodsДобавить свой ответ
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