Skorzystamy ze wzoru skróconego mnożenia na różnicę kwadratów:

${a}^{{2}}-{b}^{{2}}={\left({a}-{b}\right)}{\left({a}+{b}\right)}$ 

${a}{)}\ {f{{\left({x}\right)}}}={9}{x}^{{2}}-{81}={\left({3}{x}\right)}^{{2}}-{9}^{{2}}={\left({3}{x}-{9}\right)}{\left({3}{x}+{9}\right)}$ 

$\ \ \ {3}{x}-{9}={0}\ \ {\left|+{9}\ \ \ \vee\ \ \ {3}{x}+{9}={0}\ \ \right|}-{9}$ 

$\ \ \ {3}{x}={9}\ \ {\left|{3}\ \ \ \ \ \ \ \ \ \vee\ \ \ {3}{x}=-{9}\ \ {\left|{3}\right.}\right.}$ 

$\ \ \ {x}={3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vee\ \ \ {x}=-{3}$ 

${b}{)}\ {f{{\left({x}\right)}}}=-\frac{{1}}{{2}}{x}^{{2}}+{4}={4}-\frac{{1}}{{2}}{x}^{{2}}={2}^{{2}}-{\left(\sqrt{{\frac{{1}}{{2}}}}{x}\right)}^{{2}}=$  ${2}^{{2}}-{\left(\frac{{1}}{\sqrt{{2}}}{x}\right)}^{{2}}=$