Question

пж задание на картинке только первое​

Answer 1

$\displaystyle\bf\\\frac{16x^{2} -4y^{2} }{6x-3y} =\frac{4\cdot(4x^{2} -y^{2} )}{3\cdot(2x-y)} =\frac{4\cdot(2x -y )\cdot(2x+y)}{3\cdot(2x-y)} =\\\\\\=\frac{4\cdot(2x+y)}{3} \\\\\\x=2,5 \ \ ; \ \ y=-2\\\\\\\frac{4\cdot(2x+y)}{3}=\frac{4\cdot(2\cdot 2,5-2)}{3}=\frac{4\cdot 3}{3}=4$