$1,\dfrac{a}{b}=$ $\dfrac{b}{c}$
$⇒ac=b^{2}$
$⇒(\dfrac{a^{2}+b^{2}}{b^{2}+c^{2}})=$ $\dfrac{a^{2}+ac}{ac+c^{2}}=$ $\dfrac{a(a+c)}{c(a+c)}=$ $\dfrac{a}{c}$
$\text{ 2, đặt :}$ $\dfrac{x}{1998}=$ $\dfrac{y}{1999}=$ $\dfrac{z}{2000}=k$
$\text{⇒ x = 1998k , y = 1999k , z = 2000k}$
$(x-z)^{3}=(2000k-1998k)$$^{3}=$ $8k^{3}$
$8(x-y)^{2}(y-z)=8(1999k-1998k)$$^{2}.(1999k.2000k)$
$=8.k^{2}.k=$ $8k^{3}(đpcm)$