(sin20*sin80)*(sin40*sin60)=1/2(cos20-cos100)*1/2*(cos20-cos100)=
=1/4(cos20-cos100)²=1/4*(2sin60sin40)=1/4*4*3/4sin40=3/4*sin40
1)
x^2-4x+3=0
D=b^2-4ac= (-4)^2-4•1•3=16-12=4, 4>0
x1=(-b+√D) /2a = (4+2)/2=6/2=3
x2=(-b-√D) /2a = (4-2)/2=2/2=1
ответ: 3;1
(1-6x+9x^2)-(3x^2-16)=1-6x+9x^2-3x^2+16=6x^2-6x+17
2 cos 60°- tg45°- sin270° = 2*(1/2) - 1 - sin(180° + <span>90°) =</span><span> 1 - 1 + 1 = 1</span>