X^3<span>+4x^2-7x-10=0
x(x^2+4x-7)-10=0
x-10=0 x^2+4x-7=0
x=10 D=k-ac=2^2-1*(-7)=4+7=11
x1=-2+</span>√11
x2=-2-√11
Tg A=AC/BC=0.25 ; 16/BC=0.25 ; BC=64
ОДЗ:
x²-49≠0 (x-7)*(x+7)≠0 x₁≠-7 x₂≠7
56-x-x²≥0 |×(-1) x²+x-56≤0 D=225 √D=15
x₁=7 x₂=-8
(x-7)(x+8)≤0
-∞_____+______-8_____-_____7_____+_____+∞ ⇒
x∈[-8;7] ⇒
Ответ: x∈[-8;-7)U(-7;7).