Розв'язання завдання додаю
1
2sin4xcosx+2sin4xcos4x=0
2sin4x(cosx+cos4x)=0
2sin4x*2cos(5x/2)cos(3x)/2=0
sin4x=0⇒4x=πn,n∈z⇒x=πn/4,π∈z
cos(5x/2)=0⇒5x/2=π/2+πk,k∈z⇒x=π/5+2πk/5,k∈z
cosx/2=0⇒x/2=π/2+πm.m∈z⇒x=π+2πm,m∈z
Ответ х=-2π
2
0<2пk/7 <4π
0<2k<28
k=1;2;3;4;5;6;.....;27
0<2πk/3<4
0<2k<12
k=1;2;3;....;11
Всего 17+11=28 решений
А) -2(3а-b)+6a=-2×3a-2b+6a=6a-2b+6a=12a-2b
б) (m-2n)^2+4mn=(m-2n)(m-2n)+4=4n×2m+4
в) (x-6y)(x+6y)+9(2y)^2=2x+(6-6)+9×4y=2x+36y
<span>4х^2-10х-5=0
D= </span> (-10)2<span> - 4*4*(-5) = 100 + 80 = 180
</span>x1 = <span><span>10 - √180/</span>2*4</span> = 1.25 - 0.75√5<span> ≈ -0.42
</span>x2 = <span><span>10 + √180/</span>2*4</span> = 1.25 + 0.75√5<span> ≈ 2.92
</span>
<span>4) 2,5</span>