Question

Система х^2-х<0 -(х^2-х)<2

Answer 1

решение второго неравенства:

x²-x<2

x²-x-2<0

x²-x-2=0

D=1+8=9

x1=(1+3)/2=2

x1=(1-3)/2=-1

Получились три интервала (-∞;-1)∪(-1;2)∪(2;+∞)

Функция меньше 0 на интервале (-1;2)

решение первого неравенства

x²-x>0

x²-x=0

x(x-1)=0

x=0 и x=1

Получили три интервала (-∞;0)∪(0;1)∪(1;+∞)

Функция больше 0 на интервалах (-∞;0) и (1;+∞)

Пересечением решений первого и второго неравенств являются интервалы (-1;0)∪(1;2)

Answer 2

To solve the inequalities, let's break it down step by step:

1. $x^2 - x < 0$
2. $-(x^2 - x) < 2$

Step 1: Solve $x^2 - x < 0$

To find the solutions for this inequality, we'll first factor the expression:

$x^2 - x < 0$

Now, let's factor it: $x(x - 1) < 0$

Next, we'll find the critical points by setting each factor to zero and solving for $x$:

$x = 0$ and $x - 1 = 0$, which gives $x = 1$

Now, we can use a sign chart or test intervals to find where the inequality is true. We'll test three intervals: $(- \infty, 0)$, $(0, 1)$, and $(1, \infty)$.

1. Test $x = -1$ in $x(x - 1) < 0$: $(-1)(-1 - 1) < 0$ → $(-1)(-2) < 0$ → $2 < 0$ [False]
2. Test $x = 0.5$ in $x(x - 1) < 0$: $0.5(0.5 - 1) < 0$ → $0.5(-0.5) < 0$ → $-0.25 < 0$ [True]
3. Test $x = 2$ in $x(x - 1) < 0$: $2(2 - 1) < 0$ → $2(1) < 0$ → $2 < 0$ [False]

So, the solution for $x^2 - x < 0$ is $x \in (0, 1)$.

Step 2: Solve $-(x^2 - x) < 2$

Let's simplify the inequality first:

$-(x^2 - x) < 2$

Distribute the negative sign inside the parentheses:

$-x^2 + x < 2$

Now, let's move all terms to the left side:

$-x^2 + x - 2 < 0$

At this point, the inequality is in the same form as the one in Step 1. We can follow a similar process to find the solution:

Factor: $-1(x^2 - x - 2) < 0$

Now, find the critical points by setting each factor to zero and solving for $x$:

$x^2 - x - 2 = 0$

Now, factor the quadratic: $(x - 2)(x + 1) = 0$

Solve for $x$:

$x - 2 = 0$ → $x = 2$

$x + 1 = 0$ → $x = -1$

Test intervals: $(- \infty, -1)$, $(-1, 2)$, and $(2, \infty)$.

1. Test $x = -2$ in $-x^2 + x - 2 < 0$: $-(-2)^2 + (-2) - 2 < 0$ → $2 - 2 - 2 < 0$ → $-2 < 0$ [True]
2. Test $x = 0$ in $-x^2 + x - 2 < 0$: $-0^2 + 0 - 2 < 0$ → $-2 < 0$ [True]
3. Test $x = 3$ in $-x^2 + x - 2 < 0$: $-3^2 + 3 - 2 < 0$ → $-9 + 1 < 0$