Question

1){х^2+2х-15>0 х^2≥36 2){х-6 ____ ≥0х-6≥0{ х+103){х^2+4х-5>0 х^2-2х-8<0 ​

Answer 1

$1)\; \; \left\{\begin{array}{lll}x^2+2x-15>0\\x^2\geq 36\end{array}\right\; \; \left\{\begin{array}{lll}(x+5)9x-3)>0\\(x-6)(x+6)\geq 0\end{array}\right\; \; \left\{\begin{array}{lll}x\in (-\infty ;-5)\cup (3;+\infty )\\x\in (-\infty ;-6\, ]\cup [\, 6;+\infty )\end{array}\right\\\\\\x\in (-\infty ;-6\, ]\cup [\, 6;+\infty )\; \; -\; otvet$

$2)\; \; \left\{\begin{array}{lll}\dfrac{x-6}{x+10}\geq 0\\x-6\geq 0\end{array}\right\; \; \left\{\begin{array}{lll}x\in (-\infty ;-10)\cup [\, 6;+\infty )\\x\in [\, 6;+\infty )\end{array}\right\; \; \; \; \Rightarrow \; \; x\in [\, 6;+\infty )$

$3)\; \; \left\{\begin{array}{l}x^2+4x-5>0\\x^2-2x-80\\(x-4)(x+2)$

Answer 2

Объяснение:решение смотрите во вложении. для большей наглядности  использовал метод интервалов.

Answer 3

It looks like you've provided a set of inequalities and expressions. Let's analyze each one:

1. $x^2 + 2x - 15 > 0$ and $x^2 \geq 36$:

For the first inequality $x^2 + 2x - 15 > 0$, we can find the critical points by solving $x^2 + 2x - 15 = 0$. Factoring gives $(x + 5)(x - 3) = 0$, so the critical points are $x = -5$ and $x = 3$. Test a value from each interval to determine the solution:

• For $x < -5$, pick $x = -6$, which gives $(-6)^2 + 2(-6) - 15 = 33 > 0$.
• For $-5 < x < 3$, pick $x = 0$, which gives $0^2 + 2(0) - 15 = -15 < 0$.
• For $x > 3$, pick $x = 4$, which gives $4^2 + 2(4) - 15 = 9 > 0$.

So, the solution to $x^2 + 2x - 15 > 0$ is $x < -5$ or $x > 3$.

For the second inequality $x^2 \geq 36$, we have $x^2 - 36 \geq 0$, which can be factored as $(x - 6)(x + 6) \geq 0$. The critical points are $x = -6$ and $x = 6$. Test a value from each interval to determine the solution:

• For $x < -6$, pick $x = -7$, which gives $(-7 - 6)(-7 + 6) = 13 \geq 0$.
• For $-6 < x < 6$, pick $x = 0$, which gives $(0 - 6)(0 + 6) = -36 < 0$.
• For $x > 6$, pick $x = 7$, which gives $(7 - 6)(7 + 6) = 13 \geq 0$.

So, the solution to $x^2 \geq 36$ is $x \leq -6$ or $x \geq 6$.

2. $x - 6 \geq 0$ and $x + 10 \leq 3$:

For the first inequality $x - 6 \geq 0$, we have $x \geq 6$.

For the second inequality $x + 10 \leq 3$, we have $x \leq -7$.

So, the solution to the system of inequalities is $x \leq -7$ (from the second inequality) and $x \geq 6$ (from the first inequality), which means there is no overlapping solution.

3. $x^2 + 4x - 5 > 0$ and $x^2 - 2x - 8 < 0$:

For the first inequality $x^2 + 4x - 5 > 0$, we can find the critical points by solving $x^2 + 4x - 5 = 0$. Factoring gives $(x + 5)(x - 1) = 0$, so the critical points are $x = -5$ and $x = 1$. Test a value from each interval to determine the solution:

• For $x < -5$, pick $x = -6$, which gives $(-6)^2 + 4(-6) - 5 = 77 > 0$