已知(x^2-4x)^2+1=2x^2-8x,则x^2\x^4-8x^2+1
问题描述：
已知(x^2-4x)^2+1=2x^2-8x,则x^2\x^4-8x^2+1
答
答：(x^2-4x)^2+1=2x^2-8x(x^2-4x)^2-2(x^2-4x)+1=0(x^2-4x-1)^2=0x^2-4x-1=0显然,x不等于0,两边同除以x：x-4-1/x=0x-1/x=4两边平方得：x^2-2+1/x^2=16x^2+1/x^2=18原式=x^2/(x^4-8x^2+1)=1/(x^2+1/x^2-8)=1/(18-8)=...