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1.计算(x+1)(x²+1)(x^4+1)(x^8+1)(x^16+1)
2.p.m为何值时,x³+px-2能被x²+mx-1整除
人气:453 ℃ 时间:2020-03-16 11:00:22
解答
原式=(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x-1)/(x-1)

       =(x^2-1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)/(x-1)
       =(x^4-1)(x^4+1)(x^8+1)(x^16+1)/(x-1)=(x^8-1)(x^8+1)(x^16+1)/(x-1)
       =(x^16-1)(x^16+1)/(x-1)=(x^32-1)/(x-1)
x^3+px-2=x^3+mx^2-x-mx^2+(p+1)x-2=x(x^2+mx-1)+(-mx^2+(p+1)x-2)
所以,-mx^2+(p+1)x-2=k(x^2+mx-1)
-2=k*(-1)  ,   k=2,   m=-2,      p+1=-4,    p=-5
所以,m=-2,p=-5
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