Решение
sin²x + sinx - 2 = 0
sinx = t, t ∈ [-1;1]
t² + t - 2 = 0
D = 1 + 4*1*2 = 9
t₁ = (- 1 - 3)/2 = - 2 ∉ [-1;1]
t₂ = (-1 + 3)/2 = 1
sinx = 1
<span> x = π/2 + 2πk, k ∈ Z</span>
Log₆(x+3)(x-2) = log₆6
(x+3)(x-2) = 6
x² - 2x +3x - 6 = 6
x² + x - 12 = 0
x₁ = 3
x₂ = - 4
5y-5z+(y-z)²=5(y-z)-(y-z)²=(y-z)(5+y-z)
( xy)² xy (√3-2)(√3+2) 3-4 -1
------------ = --------= --------------------=--------- =---------
2xy 2 2 2 2
Формула: (а-в)(а+в)=а²-в²