Дано tgα=2.
найти значение выражения: [tex]\frac{sin^{4} \alpha-cos^{4}\alpha }{sin^{6} \alpha-cos^{6}\alpha }[/tex]

[tex]\frac{sin^4(x)-cos^4(x)}{sin^6(x)-cos^6(x)} = \frac{(sin^2(x)+cos^2((x) - cos^2(x))}{(sin^2(x) - cos^2((x) + sin^2(x)cos^2(x) +cos^4(x))}[/tex]

[tex]= \frac{1}{((sin^2(x) + cos^2( -sin^2(x)cos^2(x))} = \frac{1}{1-0.25sin^2(2x)}=\frac{4}{4-sin^2(2x)}[/tex]

[tex]tg(x)=2\\tg^2(x) = 4\\1+tg^2(x) = \frac{1}{cos^2(x)}\\cos^2(x) =\frac{1}{5}\\cos(x) =\frac{1}{\sqrt{5}}\\sin(x) = \sqrt{1-cos^2(x)} =\frac{2}{\sqrt{5}}\\sin(2x) = 2sin(x)cos(x) = 2*\frac{2}{\sqrt{5}}*\frac{1}{\sqrt{5}} = \frac{4}{5}[/tex]

[tex]\frac{4}{4-sin^2(2x)} = \frac{4}{4-\frac{16}{25} }=\frac{4}{3.36} =\frac{400}{336} = \frac{25}{21}\\ answer: \frac{25}{21}[/tex]

a)9a+1/2a> _6      19/2a> _6      a> _12/19    (12/19; +beskonechnost')

b)25b+1/b< _10    (25b^2+1)/b-10< _0      25b^2-10b+1< _0

d=0 b=0.2       

x(x^2-9)-4(x^2-9)=0

(x^2-9)(x-4)=0

x=-3     x=3     x=4