1. 25200之因數中為完全平方數者,其總和為多少?
2.設a和b為x^2+4x-1=0之根,則a的立方根+b的立方根=
3.已知tan(a)和tan(b)為x^2-6x+4=0的兩根,則sin^2(a)-cos^2(b)=
謝謝老師囉
數學三問
版主: thepiano
Re: 數學三問
第 1 題
25200 = 2^4 * 3^2 * 5^2 * 7
所求 = (2^0 + 2^2 + 2^4)(3^0 + 3^2)(5^0 + 5^2)
第 2 題
a + b = -4
ab = -1
令 t = a^(1/3) + b^(1/3)
(a^2)^(1/3) + (b^2)^(1/3) = [a^(1/3) + b^(1/3)]^2 - 2(ab)^(1/3) = t^2 + 2
[a^(1/3) + b^(1/3)][(a^2)^(1/3) - (ab)^(1/3) + (b^2)^(1/3)] = a + b
t(t^2 + 2 + 1) = -4
t = -1
第 3 題
tana = 3 + √5
tanb = 3 - √5
sina = (3 + √5)/√(15 + 6√5)
cosb = 1/√(15 - 6√5)
代入直接計算 (sina)^2 - (cosb)^2 = 1/3
25200 = 2^4 * 3^2 * 5^2 * 7
所求 = (2^0 + 2^2 + 2^4)(3^0 + 3^2)(5^0 + 5^2)
第 2 題
a + b = -4
ab = -1
令 t = a^(1/3) + b^(1/3)
(a^2)^(1/3) + (b^2)^(1/3) = [a^(1/3) + b^(1/3)]^2 - 2(ab)^(1/3) = t^2 + 2
[a^(1/3) + b^(1/3)][(a^2)^(1/3) - (ab)^(1/3) + (b^2)^(1/3)] = a + b
t(t^2 + 2 + 1) = -4
t = -1
第 3 題
tana = 3 + √5
tanb = 3 - √5
sina = (3 + √5)/√(15 + 6√5)
cosb = 1/√(15 - 6√5)
代入直接計算 (sina)^2 - (cosb)^2 = 1/3