初一分式求值；已知x^2+5x+1,求((x^2-5)/(x-1)+1)*(x^3-1)/(x^2-2x)*1/x+3的值

已知不完整,x^2+5x+1=?x^2+5x+1=0 还有一道(a^2-bc)/(a+b)(a+c)+(b^2-ac)/(b+c)(b+a)+(c^2-ab)/(c+a)(c+b)((x^2-5)/(x-1)+1)*(x^3-1)/(x^2-2x)*1/x+3=(x^2+x-6)/(x-1)*(x-1)(x^2+x+1)/x(x-2)*1/(x+3)=(x^2+x+1)/x=0/x=0x^2+5x+1=0(a^2-bc)/(a+b)(a+c)+(b^2-ac)/(b+c)(b+a)+(c^2-ab)/(c+a)(c+b)=[(a^2-bc)(b+c)+(b^2-ac)(a+c)+(c^2-ab)(a+b)]/(a+b)(a+c)(b+c)=0/(a+b)(a+c)(b+c)=0 望采纳，周末愉快⋯麻烦了⋯(x+y)/xy=1/(x-y)则(x^2+y^2)/xy=?