Question

Помогите пожалуйста с алгеброй (нужно полное решение этих заданий)

Answer 1

$\displaystyle\bf\\1)\\\\f(x)=\frac{x^{3} +2}{x} \\\\\\f'(x)=\frac{(x^{3} +2)'\cdot x-(x^{3}+2)\cdot x' }{x^{2} } =\frac{3x^{2} \cdot x-(x^{3}+2) }{x^{2} } =\\\\\\=\frac{3x^{3} -x^{3}-2 }{x^{2} } =\frac{2x^{3} -2}{x^{2} } \\\\\\g(x)=\frac{6x^{2} +2}{x} \\\\\\g'(x)=\frac{(6x^{2} +2)'\cdot x-(6x^{2}+2)\cdot x' }{x^{2} } =\frac{12x\cdot x-(6x^{2} +2)}{x^{2} } =\\\\\\=\frac{12x^{2} -6x^{2} -2}{x^{2} } =\frac{6x^{2} -2}{x^{2} } \\\\\\f'(x)=g'(x)$

$\displaystyle\bf\\\frac{2x^{3} -2}{x^{2} } =\frac{6x^{2} -2}{x^{2} } \\\\\\\left \{ {{2x^{3} -2=6x^{2} -2} \atop {x\neq 0}} \right. \\\\\\\left \{ {{x^{3} -3x^{2} =0} \atop {x\neq 0}} \right.\\\\\\x^{2} \cdot(x-3)=0\\\\x=0 \ - \ ne \ podxodit\\\\x-3=0\\\\x=3\\\\Otvet \ : \ B)\\\\\\2)\\\\S(t)=t^{2} +3t-2\\\\V(t)=S'(t)=(t^{2} )'+3\cdot t'-2'=2t+3\cdot 1-0=2t+3\\\\V(t)=10\\\\10=2t+3\\\\2t=7\\\\\boxed{t=3,5}$