1題：設(shè)t=x²+2x+2,原試就是6t-2/t=6-2/t,而t=x²+2x+2=（x+1)²+1≥1,所以當t=1時分式最小值為4.
2題：（a+b+c)^3=a^3+b^3+c^3+3ab²+3ac²+3ba²+3ca²+3bc²+3cb²+6abc=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc=a^3+b^3+c^3+3ab(a+b)+3abc+3ac(a+c)+3abc+3bc(b+c)+3abc-3abc=a^3+b^3+c^3+3ab(a+b+c)+3ac(a+b+c)+3bc(a+b+c)-3abc=a^3+b^3+c^3-3abc=0,所以a^3+b^3+c^3=3abc;
原式=a^3+b^3+c^3/abc=3abc/abc=3
3題：原式=1/x-1+1/x+1+2/x+1+3/x+1=6/x+2=2(3/x+1)
4題：請說明白題目