Question

已知x^2+x-3=0，求x^7+2x6-x^5-x^4-x^3-x^2-2x的值。

Answer 1

貌似等于3 。计算如下：

= x³（x²+x）-2x5-(x²+x)+x³-2x
= 9x³-2x5-9+x³-2x
= -2x5+10x³-2x-9
= -2x[(x²+x)²-2x³-6x²]-2x-9
= -2x(9-2x³-6x²)-2x-9
= 4x4+12x³-20x-9
= 4[(x²+x)²+x³-x²]-20x-9
= 36+4x³-4x²-20x-9
= 4x(x²+x-2x)-20x+27
= 4x(3-2x)-20x+27
= -8x²+12x-20x+27
= -8（x²+x）+27
= -24+27
= 3

Answer 2

已知x^2+x-3=0，
x^2+x=3
x^7+2x6-x^5-x^4-x^3-x^2-2x的值
=x^7+x^6+x^6+x^5-2x^5-x^4-x^3-x^2-x-x
=x^5(x^2+x)+x^4(x^2+x)-2x^5-x^2(x^2+x)-(x^2+x)-x
=3x^5+3x^4-2x^5-3x^2-3-x
=x^5+x^4+2x^4+2x^3-2x^3-3x^2-3-x
=x^3(x^2+x)+2x^2(x^2+x)-2x^3-3x^2-3-x
=3x^3+6x^2-2x^3-3x^2-3-x
=x^3+3x^2-3-x
=x^3+x^2+2x^2+2x-3-3x
=x(x^2+x)+2(x^2+x)-3-3x
=3x+6-3-3x
=3

Answer 3

不知