Question

Решите системы неравенств
x-4>0
ха
Решите уравнение 92-52-4=0
1
-20-1520
х-270
2²-22-1540

Answer 1

$\left \{ {{x-4 > 0} \atop {x^2-2x-15 < 0}} \right.= > \left \{ {{x > 4} \atop {(x-5)(x+3) < 0}} \right= > x > 4, -3 < x < 5$

x∈(4;5)

$\{ {{x-2\geq0 } \atop {x^2-2x-15 < 0} \right.= > \left \{ {{x\geq 2} \atop {(x-5)(x+3) < 0}} \right. \\x\geq 2, -3 < x < 5$

x∈[2;5)

$9x^4-5x^2-4=0\\x^2=t\\9t^2-5t-4=0\\D=b^2-4ac=25+144=169\\t_1=\frac{5+13}{18}=1, x^2=1= > x_1=-1, x_2=1\\t_2=\frac{5-13}{18}=-\frac{4}{9} , x^2\neq - \frac{4}{9} .$

Ответ;-1;1