Question

((n+2)/(n^2-n-6)-n/(n^2-6+9))*(2n-6)^2

Answer 1

будем считать что там n пропущено в знаменателе:

((n+2)/(n^2-n-6)-n/(n^2-6n+9))*(2n-6)^2

1. (n+2)/(n^2-n-6)=(n+2)/((n-3)(n+2))   [n^2-n-6=0;  n1=3;n2=-2по т. виета]

2. n/(n^2-6n+9)=n/((n-3)^2)

3. (n+2)/(n^2-n-6)-n/(n^2-6n+9)=(n+2)/((n-3)(n+2))-n/((n-3)^2)=

=((n+2)(n-3)-n(n+2))/((n-3)^2(n+2))=((n+2)(n-3-n))/((n-3)^2(n+2))=(-3)/((n-3)^2)

4.((-3)/((n-3)^2))*(2n-6)^2=((-3)/((n-3)^2))*4*(n-3)^2=-3*4=-12

Отв. -12