Решаем 1 неравенство
(3/5)^(x^2+5x)>1
(3/5)^(x^2+5x)>(3/5)^0
x^2+5x<0
x(x+5)<0
___-5/////////////0________
x∈(-5;0)
решаем 2 неравенство
(1/3)^(x^2-2x-3)<27
3^(-x^2+2x+3)<3^3
-x^2+2x+3<3
-x^2+2x<0
x(-x+2)<0
/////////0________2//////////
x∈(-∞;0)U(2;+∞)
смотрим общие промежутки:
//////////-5///////////0________2///////////////
х∈(-5;0)
Answers & Comments
Verified answer
1) (3/5)^(x²+5x)>1(3/5)^(x²+5x)>(3/5)⁰
x²+5x<0
x(x+5)<0
x=0 x= -5
+ - +
------ -5 ---------- 0 ------------
\\\\\\\\\\
x∈(-5; 0)
2) (1/3)^(x²-2x-3)<27
(1/3)^(x²-2x-3)<3³
(1/3)^(x²-2x-3)<(1/3)⁻³
x² -2x-3 > -3
x² -2x-3+3>0
x²-2x>0
x(x-2)>0
x=0 x=2
+ - +
------- 0 ------------ 2 -------------
\\\\\\\\ \\\\\\\\\\\\\\\\
x∈(-∞; 0)U(2; +∞)
{x∈(-5; 0)
{x∈(-∞; 0)U(2; +∞)
x∈(-5; 0) - решение системы
Ответ: (-5; 0).