Question

Даю 10 за допомогу, там де 3х² і 12 / х+7

Answer 1

$\displaystyle\bf\\1)\\\\3x^{2} +7x=6\\\\3x^{2} +7x-6=0\\\\D=7^{2} -4\cdot 3\cdot (-6)=49+72=121=11^{2} \\\\\\x_{1} =\frac{-7-11}{2\cdot 3} =\frac{-18}{6} =-3\\\\\\x_{2} =\frac{-7+11}{2\cdot 3} =\frac{4}{6} =\frac{2}{3} \\\\\\Otvet \ : \ -3 \ , \ \frac{2}{3} \\\\2)\\\\\frac{12}{x+7} +\frac{1}{x} =1\\\\\\\frac{12}{x+7} +\frac{1}{x} -1=0\\\\\\\frac{12\cdot x+x+7-x\cdot(x+7)}{x\cdot(x+7)} =0\\\\\\\frac{12x+x+7-x^{2} -7x}{x\cdot(x+7)} =0\\\\\\\frac{x^{2} -6x-7}{x\cdot(x+7)} =0$

$\displaystyle\bf\\x\neq 0 \ \ ; \ \ x+7\neq 0 \ \ \Rightarrow \ \ x\neq -7\\\\x^{2} -6x-7=0\\\\D=(-6)^{2} -4\cdot (-7)=36+28=64=8^{2} \\\\\\x_{1} =\frac{6-8}{2}=\frac{-2}{2} =-1\\\\\\x_{2} =\frac{6+8}{2} =\frac{14}{2} =7\\\\\\Otvet \ : \ -1 \ ; \ 7$