Question

ДАЮ 40 БАЛІВ
Допоможіть вирішити завдання до тесту

Answer 1

$\displaystyle\bf\\4)\\\\-2x^{2} +3x+5=0\\\\2x^{2} -3x-5=0\\\\D=(-3)^{2} -4\cdot 2\cdot (-5)=9+40=49=7^{2} \\\\\\x_{1} =\frac{3-7}{4} =\frac{-4}{4} =-1\\\\\\x_{2} =\frac{3+7}{4} =\frac{10}{4}=2,5\\\\\\\boxed{-2x^{2} +3x+5=-2(x+1)(x-2,5)}\\\\Otvet \ : \ A\\\\\\5)\\\\\frac{x^{2} }{x-7} =\frac{49}{x-7} \\\\\\\frac{x^{2} }{x-7} -\frac{49}{x-7} =0\\\\\\\frac{x^{2}-49 }{x-7} =0\\\\\\\frac{(x-7)\cdot(x+7)}{x-7} =0\\\\\\x+7=0\\\\\\\boxed{x=-7}\\\\\\Otvet \ : \ B$