Ответ:

Квадрат двучлена:   [tex]\bf x^2\pm px+q=\Big(x\pm \dfrac{p}{2}\Big)^2-\Big(\dfrac{p}{2}\Big)^2+q[/tex]  .

Выделим полные квадраты.

[tex]1)\ \ x^2-12x+39=(x-6)^2-6^2+39=(x-6)^2+3\\\\2)\ \ x^2-4x-9=(x-2)^2-2^2-9=(x-2)^2-13\\\\3)\ 2x^2-24x+36=2\cdot (x^2-12x+18)=2\cdot ((x-6)^2-6^2+18)=2\cdot (x-6)^2-36\\\\\\4)\ \ -x^2-\dfrac{1}{4}\, x+7\dfrac{3}{5}=-(x^2+\dfrac{1}{4}\, x-7\dfrac{3}{5})=-\Big(\Big(x+\dfrac{1}{8}\Big)^2-\dfrac{1}{8^2}-\dfrac{38}{5}\Big)=\\\\\\=-\Big(x+\dfrac{1}{8}\Big)^2+\dfrac{1}{64}+\dfrac{38}{5}=-\Big(x+\dfrac{1}{8}\Big)^2+\dfrac{2437}{320}=-\Big(x+\dfrac{1}{8}\Big)^2+7\dfrac{197}{320}[/tex]

[tex]5)\ \ -3x^2-1\dfrac{2}{3}\, x+39=-3x^2-\dfrac{5}{3}\, x+39=-3(x^2+\dfrac{5}{9}\, x-13)=\\\\=-3\Big(\Big(x+\dfrac{5}{18}\Big)^2-\dfrac{25}{324}-13\Big)=-3\Big(\Big(x+\dfrac{5}{18}\Big)^2-\dfrac{4237}{324}\Big)=\\\\=-3\Big(x+\dfrac{5}{18}\Big)^2+\dfrac{4237}{108}=-3\Big(x+\dfrac{5}{18}\Big)^2+39\dfrac{25}{108}\\\\\\6)\ \ -5x^2-7x+4,5=-5(x^2+1,4x-0,9)=-5\Big(\Big(x+0,7\Big)^2-0,7^2-0,9\Big)=\\\\=-5\Big(\Big(x+0,7\Big)^2-1,39\Big)=-5\Big(x+0,7\Big)^2+6,95[/tex]