Решение
<span>2tgx/(1+tg</span>²<span>x) = cosx
2sinxcosx - cosx = 0
cosx(2sinx - 1) = 0
1) cosx = 0
x = </span>π/2 + πk, k ∈ Z
2) 2sinx - 1 = 0
2sinx = 1
sinx = 1/2
x = (-1)^n arcsin(1/2) + πn, n ∈ Z
<span>x = (-1)^n (</span>π/6)<span> + πn, n ∈ Z</span>
4х^-2/3+3х^-1/3 =4х^(2*(-1/3))+3х^(-1/3).
производим замену х^(-1/3)=t.
4t²+3t-1=0
t1=-1
t2=1/4
<em>Классический метод интервалов.</em>
![\tt \dfrac{(x^2-4)(x^2-5x-14)}{x^3+8}\geq 0 \\ \dfrac{(x-2)(x+2)(x^2-7x+2x-14)}{(x+2)(x^2-2x+4)}\geq 0\\ \dfrac{(x-2)(x(x-7)+2(x-7))}{x^2-2x+4}\geq 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq -2 \\ (x-2)(x-7)(x+2) \geq 0](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B%28x%5E2-4%29%28x%5E2-5x-14%29%7D%7Bx%5E3%2B8%7D%5Cgeq%200%20%5C%5C%20%5Cdfrac%7B%28x-2%29%28x%2B2%29%28x%5E2-7x%2B2x-14%29%7D%7B%28x%2B2%29%28x%5E2-2x%2B4%29%7D%5Cgeq%200%5C%5C%20%5Cdfrac%7B%28x-2%29%28x%28x-7%29%2B2%28x-7%29%29%7D%7Bx%5E2-2x%2B4%7D%5Cgeq%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20x%20%5Cneq%20-2%20%5C%5C%20%28x-2%29%28x-7%29%28x%2B2%29%20%5Cgeq%200)
___-___(-2)___+___{2}___-___{7}___+___
<em>x∈(-2; 2]U[7; +∞)</em>
<h3 /><h3>Ответ: -1</h3>
1)(x^2 - 1) = (14 - x^2)
x^2 - 2x + 1 = 196 - 28x + x^2
x^2 - x^2 - 2x + 28x - 196 +1=0
26x = 195
x = 7,5
2) 5x - 4 = 4 - 3(5 - 2x)
5x - 4 = 4 - 15 + 6x
5x - 6x - 4 - 4 + 15 = 0
-x + 7 = 0
-x = -7
x = 7
3) -5x = 5x - 6
-5x - 5x = -6
-10x = -6
x = 0,6