102關西高中
發表於 : 2013年 6月 20日, 08:20
題目請到 Math.Pro 下載
http://math.pro/db/thread-1654-1-1.html
第 6 題
易知 xyz ≠ 0
(x + y)(x + z) = 30 ... (1)
(x + y)(y + z) = 35 ... (2)
(y + z)(x + z) = 42 ... (3)
(1)(2)(3)
[(x + y)(y + z)(x + z)]^2 = (5 * 6 * 7)^2
(x + y)(y + z)(x + z) = ± 5 * 6 * 7
解
x + y = 5
x + z = 6
y + z = 7
x = 2,y = 3,z = 4
x + y = -5
x + z = -6
y + z = -7
x = -2,y = -3,z = -4
第 12 題
老到不能再老的題目
cosα + cosβ = 2cos[(α + β)/2]cos[(α - β)/2] = 4/5 ...... (1)
sinα + sinβ = 2sin[(α + β)/2]cos[(α - β)/2] = 3/5 ...... (2)
(2) / (1)
tan[(α + β)/2] = 3/4
tan(α + β) = 24/7
http://math.pro/db/thread-1654-1-1.html
第 6 題
易知 xyz ≠ 0
(x + y)(x + z) = 30 ... (1)
(x + y)(y + z) = 35 ... (2)
(y + z)(x + z) = 42 ... (3)
(1)(2)(3)
[(x + y)(y + z)(x + z)]^2 = (5 * 6 * 7)^2
(x + y)(y + z)(x + z) = ± 5 * 6 * 7
解
x + y = 5
x + z = 6
y + z = 7
x = 2,y = 3,z = 4
x + y = -5
x + z = -6
y + z = -7
x = -2,y = -3,z = -4
第 12 題
老到不能再老的題目
cosα + cosβ = 2cos[(α + β)/2]cos[(α - β)/2] = 4/5 ...... (1)
sinα + sinβ = 2sin[(α + β)/2]cos[(α - β)/2] = 3/5 ...... (2)
(2) / (1)
tan[(α + β)/2] = 3/4
tan(α + β) = 24/7