Question

Решите неравенство подробно
Ответ: [1 ; 2]

Answer 1

4^(x²+x-3)-0,5^(2x²-6x-2)≤0
2^(2x²+2x-6)≤2^(-2x²+6x+2)
2x²+2x-6≤-2x²+6x+2
4x²-4x-8≤0 I÷4
x²-x-2≤0  D=9
x₁=2 x₂=-1.
(x-2)(x+1)≤0
-∞_____+_____-1_____-______2______+_____+∞
x∈[-1;2].

Answer 2

$2^{2(x^2+x-3)} \leq 2^{-1(2x^2-6x-2)} \\ 2^{2x^2+2x-6} \leq 2^{-2x^2+6x+2}$⇒$2x^2+2x-6 \leq -2x^2+6x+2 \\ 4x^2-4x-8 \leq 0 \\ x^2-x-2 \leq 0$
⇒$(x- \frac{1}{2})^2 \leq \frac{1}{4}+2$
$(x- \frac{1}{2})^2 \leq \frac{9}{4}$
⇒$x- \frac{1}{2} \leq \frac{3}{2} ;x- \frac{1}{2} \geq - \frac{3}{2}$
⇔$x \geq -1 \\ x \leq 2$
x∈[-1;2]