1,
5a^2 - 2003a + 8 = 0,
8b^2 - 2003b + 5 = 0,
所以,b 不等于0.【否则,5 = 0,矛盾.】
8 - 2003(1/b) + 5(1/b)^2 = 0.
所以,a,1/b都是5x^2 - 2003x + 8 = 0的根.
2003^2 - 4*5*8 > 0.
5x^2 - 2003x + 8 = 0有2个不等的根.
由韦达定理,
当ab不等于1时,
a/b = a(1/b) = 8/5.
2,
x^2 - 2(m+3)x + m^2 - 3 = 0.
0 < 4(m+3)^2 - 4(m^2 - 3) = 4[m^2 + 6m + 9 - m^2 + 3] = 4[6m + 12] = 24(m + 2),
m > -2.
x1 + x2 = 2(m+3),
(x1)(x2) = m^2 - 3.
0 = x^2 - 2(m+3)x + m^2 - 3 = x^2 - 6x - 2mx + m^2 - 3
2mx + 3 - m^2 = x^2 - 6x.
0 = (x1)^2 - 6(x1) + 2m(x2) - 3 = 2m(x1) + 3 - m^2 + 2m(x2) - 3 = 2m[x1 + x2] - m^2 = 2m[2(m+3)] - m^2 = 4m^2 + 12m - m^2 = 3m^2 - 12m = 3m(m - 4)
m = 0或者4时,x1的平方-6x1+2mx2=3.
3,
x^2 - (2k+1)y - 4 = 0,
y = x - 2,
0 = x^2 - (2k+1)(x-2) - 4 = x^2 - (2k+1)x + 2(2k+1) - 4 = x^2 - (2k+1)x + 4k - 2.
[2k+1]^2 - 4[4k-2] = 4k^2 + 4k + 1 - 16k + 8 = 4k^2 - 12k + 9
= (2k - 3)^2 >= 0.
x = [2k+1 + 2k - 3]/2 = 2k - 1,y = x - 2 = 2k - 3.
或者
x = [2k+1 - 2k + 3]/2 = 2,y = x - 2 = 0.
因 a 与 b 是0 = x^2 - (2k+1)(x-2) - 4的不同的2个根.
所以,a,b中必有1个为2.
而c = 4,所以 b = c = 4,
a + b + c = 2 + 4 + 4 = 10.
