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(-4 . t) + (t+2 . 1+t) + (t . -3) = 0
= t²-4t+2 = 0 對不對?

那請問怎麼解t²-4t+2?






答應t = 2+/-√2

2005-05-05 10:58:26 · 2 個解答 · 發問者 changchih 7 in 科學 數學

2 個解答

t^2 - 4t + 2 =0
t^2 - 4t + 4 -2 = 0
(t-2)^2 -2 = 0
(t-2)^2 =2
t-2 = +/-√2
t = +/-√2 + 2

參考看看囉。

2005-05-05 11:01:55 · answer #1 · answered by ? 7 · 0 0

代公式ㄚ!!
公式:{-b(+-)根號[b(平方)-4ac]}/2

2005-05-05 11:01:56 · answer #2 · answered by Anonymous · 0 0

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