7. 若 $a<1$, 则 $a+\frac{1}{a-1}$ 有最 值, 为
$\because a<1, \therefore a-1<0$,

$\therefore-\left(a-1+\frac{1}{a-1}\right)=(1-a)+\frac{1}{1-a} \geqslant 2(a=0$ 时取等号 $)$,

$\therefore a-1+\frac{1}{a-1} \leqslant-2, \quad \therefore a+\frac{1}{a-1} \leqslant-1$.