Предмет: Алгебра, автор: zh37525

Докажите, что при любом у квадратный трехчлен Б. –y2+6y—12 принимает отрицательные значения;

Ответы

Автор ответа: ludmilagena

0

- (y²-6y+12)<0
- (y²- 6y +9+3)<0
- ( (y-3)² +3)<0 , так как (y-3)²>0 3>0, то неравенство верно и:
-y²+6y-12<0

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