Докажите тождество16-(x+3)(x+2)=4-(6+x)(x-1)
16-(x+3)(x+2)=16-x²-5x-6=-x²-5x+10<span>
</span>4-(6+x)(x-1)=4-x²-5x+6=-x<span>²-5x+10
</span>
-x²-5x+10=-x<span>²-5x+10 ч.т.д.</span>
<h3>956) -4x-1<0</h3><h3> -4x<1</h3><h3> x > -1/4</h3><h3>960) 10x-9>0</h3><h3> 10x>9</h3><h3> x > 9/10</h3>
1)X^2+3x=0
X(x+3)=0
X=0
X+3=o
X=-3
2)x^2-4x-5=0
D=(-4)^2-4*1*(-5)=36
x1,2 = (4+-√36) / 2 = (4+-6) / 2 = 2+-3
x1 = 2-3 = -1,
x2 = 2+3 = 5.
B1. y' = -4x^3 + 8x
-4x^3 + 8x = 0
x × (8 - 4 x^2) = 0
x1=0;
4x^2 = 8
x^2 = 2
x2 = +- sqrt (2)