B^3-27+(3b^3(1+2))/(-3b^3)=b^3-27-3=b^3-30
<span>sin2xcos3c=0
sin2x=0⇒2x=πn⇒x=πn/2
cos3x=0⇒3x=π/2+πn⇒x=π/6+πn/3</span>
= 6ac² / (a-3c)(a+3c) * a-3c / ac = 6c / a+3c = 6*(-1,4) / 3,8+3*(-1,4) = -8,4 / 3,8-4,2 = -8,4/-0,4 = 21
1.
у=1/х
а) у(-2)=1/(-2) = -0,5
б) у(4) = 1/4 = 0,25
в) у(1/3) = 1/(1/3) = 3
2. у=х²
а) у(4) =4² =16
у(3) = 3² =9
у(4)>y(3)
б) у(-3)=(-3)²=9
у(-2)=(-2)²=4
у(-3)>y(-2)
в) у(2)=2²=4
у(-5)=(-5)²=25
у(2)<y(-5)
<span><span>а) 5х+3у-2х-9у = 3x - 6y
б) 2(3а-4)+5 =6a - 8 + 5 = 6a - 3
в) 15а-(а-3)+(2а-1)</span><span>= 15a - a+3+2a-1 = 16a + 2</span></span>