(1/8 cos(x)-3tg(x))'= 1/8*(-sin(x))-3*(1/cos²(x))= -1/8 sin(x)-3/cos²(x)
ответ 4
раскроем скобки в правой части тождества
<span>Решение
</span>√2sinx*cosx=cosx
<span>√2sinx*cosx - cosx = 0
cosx*(</span>√2sinx - 1) = 0
1) cosx = 0
x₁ = π/2 + πk, k ∈ Z
2) √2sinx - 1 = 0
sinx = 1/√2
x = (-1)^n * arcsin(1/√2) + πn, n ∈ Z
x₂ = (-1)^n * (π/4)<span> + πn, n ∈ Z
</span>Ответ: x₁ = π/2 + πk, k ∈ Z ; x₂ = (-1)^n * (π/4)<span> + πn, n ∈ Z</span>
<span>1. (5+y)²+y(y-7) = 25+10y+y²+y²-7y=2y²+3y+25
2. a(4-a)+(4-a)² = 4a-a²+16-8a+a²=-4a+16
3. (x-8)²-2x(6-x)² = x²-16x+64-72x+24x²-2x³=-2x³+25x²-88x+64
4. (c+7)·c-(1-c)² = c²+7c-1+2c-c²=9c-1
5. 2(a-b)² = 2a²-4ab+2b²
6. a(1+2a)² = a+4a²+4a³
7. -6(2x-y)² =-24x²+24xy-6y²
8. -y(3x-y)² = -9x²y+6xy²-y³
</span>