Question

1) (x - 1) (x + 1) < или равно0.

Answer 1

Ответ: (x²-1²)

Пошаговое объяснение:

Answer 2

Поставь вместо а=Х,у=1

будет:

${x}^{2} - 1$

Answer 3

To solve the inequality $(x - 1)(x + 1) \leq 0$, we need to find the values of $x$ that satisfy the inequality.

To do this, we first find the critical points by setting the expression $x - 1$ and $x + 1$ to zero:

$x - 1 = 0 \Rightarrow x = 1$

$x + 1 = 0 \Rightarrow x = -1$

Now, we know that the expression $(x - 1)(x + 1)$ will change sign at these critical points. Therefore, we can divide the number line into three intervals: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. We will test a point within each interval to determine the sign of the expression within that interval.

For $x = -2$, $(x - 1)(x + 1) = (-2 - 1)(-2 + 1) = (-3)(-1) = 3 > 0$ (positive) For $x = 0$, $(x - 1)(x + 1) = (0 - 1)(0 + 1) = (-1)(1) = -1 < 0$ (negative) For $x = 2$, $(x - 1)(x + 1) = (2 - 1)(2 + 1) = (1)(3) = 3 > 0$ (positive)

Now we can summarize the signs of the expression in each interval:

$(-\infty, -1)$: Positive $(-1, 1)$: Negative $(1, \infty)$: Positive

The inequality $(x - 1)(x + 1) \leq 0$ is satisfied for $x$ in the interval $(-1, 1]$ (inclusive), as the expression is either zero or negative in this range.

So, the solution to the inequality is $x \in [-1, 1]$.