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Katya132013
@Katya132013
July 2022
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Если точка (2, 3) лежит на оси параболы у=х^2+2ax-a+1, то точка (3; 0) лежит на этой параболе
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sestemika
3 = 4 + 4a - a+ 1
3a = -2
a = -2/3
y = x^2 +2ax - a + 1
y = x^2 + 2ax + 2/3 + 1
0
9 + 2*(-2/3)*3 + 5/3 ;
Нет, не лежит
2 votes
Thanks 2
Katya132013
Спасибо вам))
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Answers & Comments
3a = -2
a = -2/3
y = x^2 +2ax - a + 1
y = x^2 + 2ax + 2/3 + 1
0 9 + 2*(-2/3)*3 + 5/3 ;
Нет, не лежит