Question

已知x²+3x-1=0,求x³+5x²+5x+18的值

Answer 1

x²+3x-1=0
x²+3x=1

x³+5x²+5x+18
=x³+3x²+2x²+5x+18
=x(x²+3x)+2x²+5x+18
=x+2x²+5x+18
=2x²+6x+18
=2(x²+3x)+18
=2+18
=20

Answer 2

x^2+3x-1=0
则x^2+3x=1
x^3+5x^2+5x+18
=x(x^2+3x)+2x^2+5x+18
=x+2x^2+5x+18
=2x^2+6x+18
=2(x^2+3x)+18
=2+18
=20

Answer 3

x³+5x²+5x+18
=x³+4x²+x^2+2x+3x+19-1
=x³+4x²+2x+19+(x^2+3x-1)
=x³+4x²+2x+19
=x³+4x²+2x+19-(x^2+3x-1)
=x³+3x²-x+20
=x(x^2+3x-1)+20
=0+20=20

Answer 4

结果 20
如下x²+3x-1=0两边同乘以x得x³+3x²-x=0
x³+5x²+5x+18-0=x³+5x²+5x+18-（x³+3x²-x）=2x²+6x+18=2(x²+3x)+18=2+18=20