Question

Решить неравенство : x^3+x^2-10x больше 0

Answer 1

Ответ:

x∈($\frac{-1-\sqrt{41} }{2} ,0$)∪($\frac{-1+\sqrt{41} }{2}$,+∞)

Объяснение:

$x^{3} +x^{2} -10x>0\\x(x^{2}+x-10)>0\\\\\left \{ {{x>0} \atop {x^{2} + x -10>0}}} \right. \\\\\left \{ {{x<0} \atop {x^{2} + x -10<0}}} \right.$

[x>0

[x∈(-∞, $\frac{-1-\sqrt{41} }{2}$)∪($\frac{-1+\sqrt{41} }{2}$,+∞)

[x<0

[x∈($\frac{-1+\sqrt{41} }{2},\frac{-1-\sqrt{41} }{2}$)

[x∈($\frac{-1+\sqrt{41} }{2}$,+∞)

[x∈($\frac{-1-\sqrt{41} }{2}$,0)

x∈($\frac{-1-\sqrt{41} }{2} ,0$)∪($\frac{-1+\sqrt{41} }{2}$,+∞)

Answer 2

Ответ во вложении...