√2cos²x+cosx=0
cosx(√2cosx-+1)=0
cosx=0⇒x=π/2+πn,n∈z
-5π/2≤π/2+πn≤-π
-5≤1+2n≤-2
-6≤2n≤-3
-3≤n≤-1,5
n=-3⇒x=π/2-3π=-5π/2
n=-2⇒x=π/2-2π=-3π/2
cosx=-1/√2⇒x=-2π/3+2πk,k∈z U x=2π/3+2πm,m∈z
-5π/2≤-2π/3+2πk≤-π
-15≤-4+12k≤-6
-11≤12k≤-2
-11/12≤k≤-1/6
нет решения
-5π/2≤2π/3+2πm≤-π
-15≤4+12m≤-6
-19≤12m≤-10
-19/12≤m≤-5/6
m=-1⇒x=2π/3-2π=-4π/3
1)
3x^2 - 4 = 1/27 3x^2 = 109/27 x=√109/(3*3*9) = 1/9 √109 109 - простое число
2)
x^2 * ln(x) = e^2 * ln(e) = e^2
Вот решён наверное тебе будет все ясно там.
2) (x²+4y²)/(x-2y)+4xy/(2y-x)=(x²+4y²)/(x-2y)-4xy/(x-2y)=(x²-4xy+4y²)/(x-2y)= =(x-2y)²/(x-2y)=<span>x-2y;
4) (9a</span>²+4y²)/(3a-2y)+12ay/(2y-3a)=(9a²+4y²)/(3a-2y)-12ay/(3a-2y)=(9a²-2ay+4y²)/(3a-2y)=(3a-2y)²/(3a-2y)=<span>(3a-2y)$
6) (a</span>²x²+25y²)/(ax-5y)+10axy/(5y-ax)= (a²x²+25y²)/(ax-5y)-10axy/(ax-5y)= =(a²x²-10axy+25y²)/(ax-5y)=(ax-5y)²/(ax-5y)=<span>(ax-5y)</span>