(3-4cos2a+2cos²2a-1)/(3+4cos2a+2cos²2a-1)=
=(2cos²2a-4cos2a+2)/2cos²2a+4cos2a+2)=
=2(cos²2a-2cos2a+1)/2(cos²2a+2cos2a+1)=(cos2a-1)²/(cos2²+1)²=
=(-2sin²a)²/(2cos²a)²=4sin^4a/4cos^4a=tg^4a
F(x)=x³-2x²+1 x₀=2
yk=y(x₀)+y`(x₀)*(x-x₀)
y(2)=2³-2*2²+1=8-2*4+1=8-8+1=1
y`(2)=3*x²-2*2*x=3x²-4x=3*2²-4*2=3*4-8=12-8=4 ⇒
yk=1+4*(x-2)=1+4x-8=4x-7.
Ответ: yk=4x-7.
(√32-3)²=(√32)²-2*3*√32+3²=32-6*√(16*2)+9=41-6*4√2=41-24√2