[Solution] Notice that $\frac{x}{x-1}=\frac{y^{2}+2 y-1}{\left(y^{2}+2 y-1\right)-1}$, by checking the answer, we can immediately determine that $\quad x=y^{2}+2 y-1$. Therefore, the answer is $(A)$.
$1 \cdot 22$ If $\frac{x}{x-1}=\frac{y^{2}+2 y-1}{y^{2}+2 y-2}$, then $x$ equals
(A) $y^{2}+2 y-1$.
(B) $y^{2}+2 y-2$.
(C) $y^{2}+2 y+2$.
(D) $y^{2}+2 y+1$.
(E) $-y^{2}-2 y+1$.
(32nd American High School Mathematics Examination, 1981)