初三奥数题

Question

已知X(三次方)-3√2X(平方)+6X-2√2-8=0
则X(五次方)-41X(平方)+1=?
答案是-13
告诉我过程!
谢谢各位大侠了
我分不多.
见谅!

Answer 1

X^3-3√2X^2+6X-2√2-8=0
x^3-3√2x^2+3*(√2)^2x-(√2)^3-8=0
利用完全立方公式得：
(x-√2)^3=8

或者：
(x^3-2√2x^2+2x)-√2x^2+4x-2√2-8=0
x(x-√2)^2-(√2x^2-4x+2√2)-8=0
x(x-√2)^2-(x-√2)(√2x-2)-8=0
(x-√2)(x^2-√2x-√2x+2)-8=0
(x-√2)(x^2-2√2x+2)-8=0
(x-√2)(x-√2)^2-8=0
(x-√2)^3=8

x-√2=2
x=2+√2

x^5-41x^2+1
=x^2(x^3-41)+1
=(2+√2)^2*[(2+√2)^3-41]+1
=(4+4√2+2)(8+12√2+12+2√2-41)+1
=(6+4√2)(14√2-21)+1
=2*(3+2√2)*7(2√2-3)+1
=14*[(2√2)^2-3^2]+1
=14*(8-9)+1
=-14+1
=-13