Question

Помогите пож-та!
Задание во вложении!

Answer 1

[2log²(3)2-(2+log(3)2)²-log(3)2(2+log(3)2)]/(2log(3)2+log(3)2+2)=
=[2log²(3)2-4-4log(3)2-log²(3)2-2log(3)2-log²(3)2]/(2+3log(3)2)=
=(-4-6log(3)2)/(2+3log(3)2=-2(2+3log(3)2)/(2+3log(3)2=-2