Question

(a-2)(a-4)+(b+1)(b-5)>b-10

Answer 1

$a^{2} -4a-2a+8$+$b^{2}-5b+b-5 \ \textgreater \ b-10$
$a^{2}-6a+ 8+b^{2} -4b-5-b+10\ \textgreater \ 0 \\ a^{2} + b^{2} -6a-5b+13\ \textgreater \ 0$
Din pacate numai pina aici.

Answer 2

(a-2)(a-4)+(b+1)(b-5)>b-10
a²-4a-2a+8+b²-5a+1-5>b-10
a²+b²+4>b-10
a²+b²+b>7