(5x²-3x+1)(x²-4x-12)<0
5x²-3x+1=0
D=9-4*5=-11<0 значит
5x²-3x+1>0 при любых значениях х.
х²-4х-12=0
В=16+4*12=64=8²
х₁=(4+8)/2=6
х₂=(4-8)/2=-2
<span>(5x²-3x+1)(x-6)(x+2)<0
</span>x∈(-2; 6)
<span>√0,36*2,5=0.6*2.5=1.5</span>
sin^2a+cos^2a/tg^2a*cos^2a - cos^2a/1-cos^2a =
1/(sin^2a/cos^2a *cos^2a)- cos^2a/sin^2a =
1/sin^2a - cos^2a/sin^2a =(1 - cos^2a)/sin^2a=
sin^2a/sin^2a =1