9*loq7(x^2+x-2)<=10*loq7(7)+loq7((x+2)^(-1)*(x-1)^9)
ОДЗ хЄ(-§; -2)+(1; +§)
loq7(7^10)+loq7((x+2)^(-1)*(x-1)^9*)-loq7(((x+2)*(x-1))^9)>=0
loq7((7^10*(x-1)^9*(x+2)^(-1))/((x+2)^9*(x-1)^9)>=0
loq7(7^10*(x+2)^(-10))>=loq7(1)
7^10*(x+2)^(-10)>=1
7^10>=(x+2)^10
7>=x+2
x=<5 с учётом ОДЗ хЄ(-§; -2)+(1; 5]
10m^2+2mn-20mn-4n^2-m^2-n^2-2mn=9m^2-20mn-5n^2
Применим формулу синуса половинного угла слева и синуса двойного угла справа:
2sin²(x/2) = 2·2sin(x/2)cos(x/2)·sin(x/2)
2sin²(x/2) = 4sin²<span>(x/2)cos(x/2)
</span>2sin²(x/2) - 4sin²<span>(x/2)cos(x/2) = 0
</span>2sin²(x/2) ·(1 - 2<span>cos(x/2)) = 0
</span>sin²(x/2) = 0 или 1 - 2<span>cos(x/2) = 0
</span>x/2 = πn, n∈Z cos(x/2) = 1/2
x = 2πn, n∈Z x/2 = π/3 + 2πk, k∈Z или x/2 = - π/3 + 2πm, m∈Z
x = 2π/3 + 4πk, k∈Z x = - 2π/3 + 4πm, m∈Z<span>
</span> 2sin²(x/2) - 4sin²(x/2)cos(x/2) = 0
2sin²(x/2) - 2·2sin²<span>(x/2)cos(x/2) = 0
</span> _______ _______ это выносим
2sin²(x/2) · ( 1 - 2<span>cos(x/2)) = 0</span>
8x-4x-10=6x+4
4x-6x=4+10
-2x=14
x=-7