(x²-6x-7)/(1-1/x²)²≤0 ( x≠0, x≠1, x≠-1) [(x+1)(x-7)]/[(x²-1)/x²]²≤0
-3x+3>0 ⇔ x<1 ⇔<br>
[(x+1)(x-7)x⁴]/[(x-1)²(x+1)²]≤0 [(x-7)x⁴]/[(x-1)²(x+1)]≤0
x<1 ⇔ x<1 ⇔<br>
1) [(x-7)x⁴]/[(x-1)²(x+1)]≤0 ( x≠0, x≠1, x≠-1)
+ - - +
-----------(-1)-/////////////////(1)/////////////////(7)-------------------------------------
2) x<1<br>-////////////////////////////////////(1)---------------------------------------------------------
ответ: x∈(-1;1)