a<m<b, b-a=?

2014-11-07 7:25 am
方程式49^x-(m-7)7^x+9=0有相異兩正根, 則a<m<b, b-a=?

回答 (1)

2014-11-07 11:37 pm
✔ 最佳答案
相異兩正根 α、β (7^x)^2 – (m-7)(7^x) + 9 = 0, t = 7^x t^2 – (m-7)t + 9 = 0, 相異兩根7^α、7^β7^α7^β= 9, 7^α+ 7^β= m-7 (m-7)^2 – 36 > 0, m > 13 或 m < 1 ---(1) 7^α> 1, 7^β > 1(7^α- 1)( 7^β- 1) > 07^α7^β – (7^α+ 7^β) + 1 > 09 – (m-7) + 1 > 0 m < 17 --- (2) 由(1)、(2)知13 < m < 17, a =13, b = 17 b – a = 17 – 13 = 4


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