Question

已知x^2-2x-1=0，求2x^5+4x^4-3x^3+3x^2-x+4的值

Answer 1

x²-2x-1=0
x²=2x+1
所以
x³=x*x²
=x(2x+1)
=2x²+x
x^4=(x²)²
=(2x+1)²
=4x²+4x+1
x²-2x-1=0
所以x²-2x=1
原式=2x*(4x²+4x+1)+4(4x²+4x+1)-3(2x²+x)+3x^2-x+4
=8x^3+8x^2+2x+16x^2+16x+1-6x^2-3x+3x^2-x+4
=8(2x^2+x)+21x^2+14x+5
=16x^2+8x+21x^2+14x+5
=37x^2+22x+5
=37(2x+1)+22x+5
=74x+37+22x+5
=96x+42
解x^2-2x-1=0
解得x=1±根号2
原式=96±96根号2+42
=138±96根号2