y二次+y負二次=7 ,y>0 求下列各式值
y+1/y
(y-1/y)二次
(y二次-y負二次)二次
F3 maths 關於恆等式
2008-08-03 3:55 am
回答 (1)
2008-08-03 5:09 am
✔ 最佳答案
(1)(y + 1/y)^2 = y^2 + 2 + 1/y^2.But y^2 + 1/y^2 = 7. That is (y+1/y)^2 = 7 + 2 = 9. Therefore, y+1/y = sqrt9 = 3.
(2)(y-1/y)^2 = y^2 + 1/y^2 - 2 = 7 - 2 = 5.
(3)
y^2 - 1/y^2 = (y-1/y)(y+1/y) = sqrt5 x 3 = 3sqrt5.
Therefore, (y^2 - 1/y^2)^2 = (3sqrt5)^2 = 9 x 5 = 45.
收錄日期: 2021-04-25 22:35:24
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