Y=(√x-2)/(sin²x-sinx-2)
D(y):
1) x-2≥0, x≥2
2) sin²x-sinx-2≠0
sinx=t, t∈[-1;1]
t²-t-2=0
D=(-1)²-4*(-2)=9
t=(1+3)/2=4/2=2 - посторонний корень, т.к. t∈[-1;1]
t=(1-3)/2=-2/2=-1
sinx=-1
x=-π/2+2πn, n∈Z
Ответ: x∈[2;+∞) U x≠-π/2+2πn, n∈Z
2x^3 - 11x^2 + 17x - 6 = 2x^3 - 4x^2 - 7x^2 + 14x + 3x - 6 =
= 2x^2 * (x - 2) - 7x * (x - 2) + 3 * (x - 2) =
= (x - 2) * (2x^2 - 7x + 3) = (x - 2) * (x - 3) * (2x - 1) = 0
x1 = 2
x2 = 3
x3 = 0,5
Среднее арифметическое = (2 + 3 + 0,5) / 3 = 11/6
13^2 - 11^2 = 48
11^2 - 5^2 = 96
По теореме Виета
x₁+x₂=5
x₁*x₂=6
x₁=2 ;х₂=3