Question

2 задания по алгебре. Помогите пожалуйста

Answer 1

$f(x)=\frac{1}{x}\\\\f(x+4)=7f(x+8)\\\\\frac{1}{x+4} =7*\frac{1}{x+8}\\\\\frac{1}{x+4}-\frac{7}{x+8}=0\\\\\frac{x+8-7x-28}{(x+4)(x+8)}=0\\\\\frac{-6x-20}{(x+4)(x+8)}=0\\\\\left \{ {{-6x-20=0} \atop {x\neq-4;x\neq-8}} \right. \\\\-6x-20=0\\\\-6x=20</p><p>[tex]x=-3\frac{1}{3}\approx-3,33$

$2)f(x)=\frac{2}{x}\\\\f(6)=\frac{2}{6}=\frac{1}{3}\\\\f(-5)=\frac{2}{-5}=-\frac{2}{5}\\\\f(6)+f(-5)=\frac{1}{3}-\frac{2}{5}=\frac{5-6}{15}=-\frac{1}{15}\approx-0,07$