已知多项式2x²+3xy-2y²-x+8y-6=(x+2y+m)·(2x-y+n),那么(m^4+8)/(n²-1)=
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![已知多项式2x²+3xy-2y²-x+8y-6=(x+2y+m)·(2x-y+n),那么(m^4+8)/(n²-1)=](/uploads/image/z/3008420-44-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%A4%9A%E9%A1%B9%E5%BC%8F2x%26%23178%3B%2B3xy-2y%26%23178%3B-x%2B8y-6%3D%28x%2B2y%2Bm%29%C2%B7%282x-y%2Bn%29%2C%E9%82%A3%E4%B9%88%EF%BC%88m%5E4%2B8%29%2F%28n%26%23178%3B-1%29%3D)
已知多项式2x²+3xy-2y²-x+8y-6=(x+2y+m)·(2x-y+n),那么(m^4+8)/(n²-1)=
已知多项式2x²+3xy-2y²-x+8y-6=(x+2y+m)·(2x-y+n),那么(m^4+8)/(n²-1)=
已知多项式2x²+3xy-2y²-x+8y-6=(x+2y+m)·(2x-y+n),那么(m^4+8)/(n²-1)=
2x²+3xy-2y²-x+8y-6=(x+2y+m)(2x-y+n)
=2x²-xy+nx+4xy-2y²+2ny+2mx-my+mn
=2x²+3xy-2y²+(n+2m)x+(2n-m)y+mn
所以n+2m=-1,2n-m=8,所以m=-2,n=3
所以原式=[(-2)^4+8]/(9-1)
=(16+8)/8
=3