已知3x²-x-1=0,求6x³+7x²-5x+1999的值

已知3x²-x-1=0,求6x³+7x²-5x+1999的值
已知3x²-x-1=0,求6x³+7x²-5x+1999的值

已知3x²-x-1=0,求6x³+7x²-5x+1999的值
6x³+7x²-5x+1999
=x(2*3x²+7x-5)+1999
=x(2*3x²-2x-2+9x-3)+1999
=x[2(3x²-x-1)+9x-3]+1999
=x[0+9x-3]+1999
=9x²-3x+1999
=3(3x²-x-1)+2002
=2002

3x²-x-1=0
3x² = x+1
则
6x³+7x²-5x+1999
= 2x(x+1) + 7/3 * (x+) - 5x + 1999
= 2x² + 2x + 7x/3 + 7/3 - 5x + 1999
= 2x² - 2x/3 + 1/3 + 2001
= 2/3 * (x+1) - 2x/3 + 1/3 + 2001
= 2x/3 + 2/3 - 2x/3 + 1/3 + 2001
= 2002

2002这种题直接后者除以前者，然后得余数