Question

已知 x=4-√3,则分式(x^4-6^3-2^2+18x+23)/(x^2-8x+15)的值为

（x^4-6x^3-2x^2+18x+23）/(x^2-8x+15)

Answer 1

解：∵当x=4-√3时，x²-8x+15=（x-3）(x-5）=（1-√3）(-1-√3)=2
x^4-6x³-2x²+18x+23
=x²（x²-8x+15）+2x（x²-8x+15）-（x²-8x+15）-20x+38
=2x²+4x-2-20x+38
=2x²-16x+36
=2（x²-8x+15）+6
=4+6
=10，
∴（x^4-6x³-2x²+18x+23）/（x²-8x+15）=10/2=5
故答案为：5

Answer 2

x=4-√3
x-4=-√3
(x-4)²=3
x²-8x+16-3=0
x²-8x+13=0

x^4-6x³-2x²+18x+23

=x^4-8x³+13x²+2x³-15x²+18x+23
=x²(x²-8x+13)+2x³-16x²+26x+x²-8x+23
=x²(x²-8x+13)+2x(x²-8x+13)+x²-8x+23
=x²-8x+23

因为x²-8x+13=0
x²-8x+13+10=10
x²-8x+13+2=2
（x^4-6x^3-2x^2+18x+23）/(x^2-8x+15)=10/2=5

Answer 3

我算的是5
不知道对不对

Answer 4

你的这个分式有误吧？分子怎么出现三个常数项？

追问

不好意思，打错了、分子x^4-6x^3-2x^2+18x+23