x=√2-1,x+1-√2=0,(x+1-√2)(x+1+√2)=0,x^2+2x-1=0. x^5+2x^4+x^3+4x^2-x+1 =(x^5+2x^4-x^3)+(2x^3+4x^2-2x)+(x+1) =x^3(x^2+2x-1)+2x(x^2+2x-1)+(x+1) =x+1=√2-1+1=√2.
解： x=(根2)-1→(x+1)^2=2→x^2+2x-1=0. 故x^5+2x^4+x^3+4x^2-x+1 =x^3(x^2+2x-1)+2x(x^2+2x-1)+(x+1) =(x^3+2x)(x^2+2x-1)+(x+1) =(x^3+2x)×0+(x+1) =x+1 =根2.
x^5+2x^4+x^3+4x^2-x+1 =x³(x²+2x+1)+4x²-x+1 =x³(x+1)²+4x²-x+1 =x³(√2-1+1)²+4x²-x+1 =2x³+4x²-x+1 =2x（x²+2x+1）-3x+1 =2x(x+1)²-3x+1 =2x(√2-1+1)²-3x+1 =4x-3x+1 =x+1 =√2
答：这道题的关键是降系, x^4+4x^3-4x-4 首先我们要对它化简,最好能因式分解,仔细观察,可以看出,这里面少了2次方,我们补了一个x^2后,应该就能分解...详情>>
答：详情>>
