已知x1+x1的3次方=3…x2+三次根号下x2=3……求x1+x2的值

已知x1+x1的3次方=3…x2+三次根号下x2=3……求x1+x2的值

设t = (x2)^(1/3),
则
t + t^3 = 3
所以,x1,(x2)^(1/3)都是3次方程
x + x^3 = 3,
的根.
设f(x) = x^3 + x - 3,
f'(x) = 3x^2 + 1 > 0.
所以f(x) = x^3 + x - 3是单调递增函数.
f(x)至多只能有1个零点.
而f(0) = -3 < 0,
f(3) = 27 > 0.
所以f(x)在区间（0,3）上至少有1个零点.
这样,
f(x)有且只有1个零点.
所以,
x1 = (x2)^(1/3)
x1 + x2 = (x2)^3 + x2 = 3.
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随便练习一哈3次方程的求根~~~
x^3 + x - 3 = 0.
令x = u + v
0 = (u+v)^3 + (u+v) - 3
= u^3 + v^3 + 3uv(u+v) + (u+v) - 3
= (u+v)[3uv + 1] + [u^3 + v^3 - 3]
这样,
只要能找到 u,v,满足
uv = -1/3,
u^3 + v^3 = 3
则x = u + v一定是x^3 + x - 3 = 0的根.
也就是说,如果能找到a,b,满足
a = u^3, b = v^3
a + b = u^3 + v^3 = 3,
ab = (uv)^3 = -(1/3)^3
则,x = u + v = a^(1/3) + b^(1/3)一定是x^3 + x - 3 = 0的根.
而由韦达定理.
a,b是方程z^2 - 3z - (1/3)^3= 0的2个根.
所以,
a = {3 + [9 + 4(1/3)^3 ]^(1/2)}/2
b = {3 - [9 + 4(1/3)^3 ]^(1/2)}/2
x = a^(1/3) + b^(1/3)
= ({3 + [9 + 4(1/3)^(1/3) ]^(1/2)}/2)^(1/3) +
+ ({3 - [9 + 4(1/3)^(1/3) ]^(1/2)}/2)^(1/3)
是x^3 + x - 3 = 0的1个根.
又因为
e^[i2PI] = 1
e^[i2PI/3], e^[-i2PI/3] 和 1 = e^[i2PI]是1的3个3次方根
【 [e^(i2PI/3)]^3 = [e^(-i2PI/3)]^3 = 1^3 = 1,
[e^(i2PI/3)]^2 = e^(-i2PI/3),
[e^(-i2PI/3)]^2 = e^(i2PI/3) 】
所以,
[a^(1/3)e^(i2PI/3)]^3 + [b^(1/3)e^(-i2PI/3)]^3 = a + b = 3
a^(1/3)e^(i2PI/3)*b^(1/3)e^(-i2PI/3) = (ab)^(1/3) = -1/3
因此,x = a^(1/3)e^(i2PI/3) + b^(1/3)e^(-i2PI/3)也是x^3 + x - 3 = 0的根.
同样,
[a^(1/3)e^(-i2PI/3)]^3 + [b^(1/3)e^(i2PI/3)]^3 = a + b = 3
a^(1/3)e^(-i2PI/3)*b^(1/3)e^(i2PI/3) = (ab)^(1/3) = -1/3
因此,x = a^(1/3)e^(-i2PI/3) + b^(1/3)e^(i2PI/3)也是x^3 + x - 3 = 0的根.
这样,
x^3 + x - 3 = 0的3个根就找到了,分别是,
a^(1/3) + b^(1/3),
a^(1/3)e^(i2PI/3) + b^(1/3)e^(-i2PI/3)和
a^(1/3)e^(-i2PI/3) + b^(1/3)e^(i2PI/3).
其中,
a = {3 + [9 + 4(1/3)^3 ]^(1/2)}/2
b = {3 - [9 + 4(1/3)^3 ]^(1/2)}/2