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x^3+3x%4如何分解为(x%1)(x^2+x+4)?

由于 (1-x)(1+x+x^2)=1-x^3 1+x+x^2+x^3+x^4=(1-x^3)/(1-x)+x^3*(1+x) =(1-x^3)/(1-x)+[x^3*(1+x)*(1-x)]/(1-x) =(1-x^3)/(1-x)+[x^3*(1-x^2)]/(1-x) =(1-x^3)/(1-x)+(x^3-x^5))/(1-x) =(1-x^3+x^3-x^5)/(1-x) =(1-x^5)/(1-x)

(x-1)(x+2)(x+3)(x+4)+24 应该是这个吧: (x+1)(x+2)(x+3)(x+4)-24 =(x+1)(x+4)(x+2)(x+3)-24 =(x²+5x+4)(x²+5x+6)-24 =(x²+5x)²+10(x²+5x)+24-24 =(x²+5x)(x²+5x+10) =x(x+5)(x ²+5x+10)

供参考。

X(X^3-X^2-X-2)=0 X(X^3-2X^2+X^2-2X+X-2)=0 X[X^2(X-2)+X(X-2)+(X-2)]=0 X(X-2)(X^2+X+1)=0

x^4-x^3+3x^2-2x+2 =(x^4+3x^2+2)-(x^3+2x) =(x^2+2)(x^2+1)-(x^2+2)x =(x^2+2)(x^2-x+1)

把(X²+X+2)当成整体。 原式=(X²+X+2-1)(X²+X+2+1)²+4 =(X²+X+2)²-1+4 =(X²+X+2)²+3 无法分解。 请检查题目并追问。

(x+y)^2-4(x+y-1) =(x+y)^2-4[(x+y)-1] =(x+y)^2-4(x+y)+4 =(x+y-2)^2

原式x^4+2x^3+3x^2+2x+1 =x^4+2x^2+1+2x^3+x^2+2x =(x^2+1)^2+x^3+x+x^3+x^2+x =(x^2+1)^2+x(x^2+1)+x(x^2+x+1) =(x^2+1)(x^2+x+1)+x(x^2+x+1) =(x^2+1+x)(x^2+1+x) =(x^2+1+x)^2

令t=x+4, 则x=t-4, 将f(x)展开成t 幂级数即可。 f(x)=1/(x+1)(x+2)=1/(x+1)-1/(x+2) =1/(t-3)-1/(t-2) =-1/(1-t/3)+1/(1-t/2) =-[1+t/3+t²/3²+...]+[1+t/2+t²/2²+.....], 收敛域为|t|

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