............................
X >= -4/5
x >= 2
2) D = b^2 - 4ac = (-4)^2<span> - 4·1·3 = 16 - 12 = 4
x1 = 1;
x2 = 3;</span>
a³-(a-3)³=54+9a²
a³-(a-3)³=27+27+9a²
(a³-27)-(a-3)³=9a²+27
(a³-3³)-(a-3)³=9a²+27
(a-3)(a²+3a+9)-(a-3)³=9a²+27
(a-3)(a²+3a+9)-(a-3)²)=9a²+27
(a-3)(a²+3a+9-(a²-6a+9)=9a²+27
(a-3)(a²+3a+9-a²+6a-9)=9a²+27
(a-3)*9a=9a²+27
9a²-27a=9a²+27
-27a=27 |÷(-27)
a=-1.
Ответ: a=-1.
F(x)=∫ 4sin(2x+π/4) dx= -4*1/2*cos(2x+π/4)+C=-2*cos(2x+π/4)+C