Question

Решите показательное уравнение и неравенство:
(5/3)^х-2=(27/125)^4+x²;

10^3x+1>0,001

Answer 1

$left(frac53right)^{x-2}=left(frac{27}{125}right)^{4+x^2}\left(frac35right)^{2-x}=left(frac35right)^{3(4+x^2)}\2-x=12+3x^2\3x^2+x+10=0\D textless 0;-;kopHeu;HET.\\10^{3x+1} textgreater 0,001\10^{3x+1} textgreater 10^{-3}\3x+1 textgreater -3\3x textgreater -4\x textgreater -1frac13$

Answer 2

(5/3)^(х-2)=(3/5)^3*(4+х²)
(3/5)^(2-х)=(3/5)^(12+3х²)
2-х=12+3х²
3х²+х+12-2=0
3х²+х+10=0
Д=1²-4*3*10=1-120=-119<0
нет корней
10^(3х+1)>0,001
10^(3х+1)>10^(-3)
3х+1>-3
3х>-3-1
3х>-4
х>-4/3=-1.1/3