✔ 最佳答案
(1)
4^x + 6^x = 9^x
4^x + 6^x - 9^x = 0
(2^x)^2 + (2^x)(3^x) - (3^x)^2 = 0
[(2/3)^x]^2 + (2/3)^x - 1 = 0
(2/3)^x = (-1 + root5) / 2
x log(2/3) = log[(-1 + root 5) / 2]
x = [log(root 5 - 1) - log2] / (log 2 - log 3)
(2)
(x - a)(x - b)(x - r) = 0
x^3 - (a + b + r)x^2 + (ab + br + ra)x - abr = 0
so, a + b + r = 6, ab + br + ra = 3, abr = -1
a^3 + b^3 + r^3
= (a + b + r)^3 - 3(ab^2 + br^2 + ra^2 + ar^2 + ba^2 + rb^2) - 6abr
= (a + b + r)^3 - 3[(a + b + r)(ab + br + ra) - 3abr] - 6abr
= (a + b + r)^3 - 3(a + b + r)(ab + br + ra) + 3abr
= (6)^3 - 3(6)(3) + 3(-1) = 159
(a^2 + 1)(b^2 + 1)(r^2 + 1)
= (abr)^2 + (a^2)(b^2) + (b^2)(r^2) + (r^2)(a^2) + a^2 + b^2 + r^2 + 1
= (abr)^2 + (ab + br + ra)^2 - 2(arb^2 + bar^2 + rba^2)
+ (a + b + r)^2 - 2(ab + br + ra) + 1
= (abr)^2 + (ab + br + ra)^2 - 2abr(a + b + r) + (a + b + r)^2 - 2(ab + br + ra) + 1
= (-1)^2 + (3)^2 - 2(-1)(6) + (6)^2 - 2(3) + 1 = 53
2014-06-09 15:08:06 補充:
第三題有待修正
2014-06-09 15:09:59 補充:
原來貓大已答了,sorry