Question

Помогите решить неравенство, дохожу до замены, а дальше никак.

Answer 1

$\frac{5^x+1}{0,2-5^x}\geq2log_2\sqrt{2}\\\\\frac{5^x+1}{0,2-5^x}\geq2*\frac{1}{2}\\\\\frac{5^x+1}{0,2-5^x}\geq1\\\\y=5^x\\\\\frac{y+1}{0,2-y}\geq1\\\\\frac{y+1}{0,2-y}-1\geq0\\\\\frac{y+1-(0,2-y)}{0,2-y}\geq0\\\\\frac{y+1-0,2+y}{0,2-y}\geq0\\\\\frac{2y+0,8}{0,2-y}\geq0\\\\\frac{2(y+0,4)}{0,2-y}\geq0$

_______[-0,4]//////////(0,2)_______

$-0,4\leq y<0,2\\\\-0,4\leq 5^x<5^{-1}\\\\5^x\geq -0,4\\5^x>0\\x\in R\\\\5^x<0,2\\5^x<5^{-1}\\5>1=>x<-1$

Ответ: (-∞;-1)