1.若x^2 - 3x -1=0,求代数式2x^3 - 3x^2 -11x +8的值．2.若3x^2 - x=1,求代数式6x^3 +7x^2 -5x +1999的值．

x^2 - 3x -1=0
x^2-3x=1
2x^3 - 3x^2 -11x +8
=2x(x^2-3x)+3x^2-11x+8
=2x+3x^2-11x+8
=3x^2-9x+8
=3(x^2-3x)+8
=3+8
=11
6x^3 +7x^2 -5x +1999
=2x(3x^2-x)+9x^2-5x+1999
=2x+9x^2-5x+1999
=9x^2-3x+1999
=3(3x^2-x)+1999
=3+1999
=2002
1 2x^3 - 3x^2 -11x +8 = (2x+3)(x^2-3x-1) +11 = 11
2 条件即3x^2 -x-1=0
而6x^3 +7x^2 -5x +1999 = (2x+3)(3x^2-x-1) +2002 =2002
因为x^2 - 3x -1=0,
所以2x^3 - 3x^2 -11x +8
=2x(x^2-3x-1)+3x^2-9x+8
=3x^2-9x+8
=3(x^2-3x-1)+11
=11;
(2)因为3x^2-x=1,
所以3x^2-x-1=0,
所以6x^3 +7x^2 -5x +1999
=2x(3x^2 - x-1)+9x^2-3x+1999
=9x^2-3x+1999
=3(3x^2-x-1)+2002
=2002.

第一种解法:x^2 - 3x -1=0
x^2-3x=1
2x^3 - 3x^2 -11x +8
=2x(x^2-3x)+3x^2-11x+8
=2x+3x^2-11x+8
=3x^2-9x+8
=3(x^2-3x)+8
=3+8
=11
第二种解法:6x^3 +7x^2 -5x +1999
=2x(3x^2-x)+9x^2-5x+1999
=2x+9x^2-5x+1999
=9x^2-3x+1999
=3(3x^2-x)+1999
=3+1999
=2002

1 2x^3 - 3x^2 -11x +8 = (2x+3)(x^2-3x-1) +11 = 11
2 条件即3x^2 -x-1=0
而6x^3 +7x^2 -5x +1999 = (2x+3)(3x^2-x-1) +2002 =2002

因为x^2 - 3x -1=0,
所以2x^3 - 3x^2 -11x +8
=2x(x^2-3x-1)+3x^2-9x+8
=3x^2-9x+8
=3(x^2-3x-1)+11
=11;
(2)因为3x^2-x=1,
所以3x^2-x-1=0,
所以6x^3 +7x^2 -5x +1999
=2x(3x^2 - x-1)+9x^2-3x+1999
=9x^2-3x+1999
=3(3x^2-x-1)+2002
=2002.

表示出X*X就可以了,代,只管代,到后来,可以约的

2002

一个11，一个2002

x^2 - 3x -1=0
x^2-3x=1
2x^3 - 3x^2 -11x +8
=2x(x^2-3x)+3x^2-11x+8
=2x+3x^2-11x+8
=3x^2-9x+8
=3(x^2-3x)+8
=3+8
=11
6x^3 +7x^2 -5x +1999
=2x(3x^2-x)+9x^2-5x+1999
=2x+9x^2-5x+1999
=9x^2-3x+1999
=3(3x^2-x)+1999
=3+1999
=2002

第一题等于11
第二题等于2002