1.(n-6)(2n+3)-(n-7)в квадрате-(2+n)(n-2)

Question

1.(n-6)(2n+3)-(n-7)в квадрате-(2+n)(n-2)

Answer 1

(n-6)(2n+3)-(n-7)²-(2+n)(n-2)=
2n²-12n-18+3n-(n²-14n+49)-(n²-4)=
2n²-12n-18+3n-n²+14n-49-n²+4=
5n-63