Question

НУЖНА ПОМОЩЬ!!!!!!!!!

Answer 1

$\displaystyle\bf\\\frac{18x^{2} -48xy+32y^{2} }{9x-12y} =\frac{2\cdot(9x^{2} -24xy+16y^{2}) }{3(3x-4y)} =\frac{2\cdot(3x-4y)^{2} }{3(3x-4y)} =\\\\\\=-\frac{2\cdot(4y-3x)^{2} }{3\cdot(4y-3x)} \\\\\\4y-3x=-0,2\\\\\\-\frac{2\cdot(4y-3x)^{2} }{3\cdot(4y-3x)} =-\frac{2\cdot(-0,2)^{2} }{3\cdot(-0,2)}=\frac{2\cdot0,04}{0,6} =\frac{0,08}{0,6} =\frac{2}{15}$