Question

Іх-2I + ІхІ = 4; б) І x-2 I + Іх+2І=4

Answer 1

Іх-2I + ІхІ = 4;

х¹=-3

І x-2 I + Іх+2І=4

х∈[-2;2]

Answer 2

It seems like you're asking to solve a system of equations involving absolute values:

a) $|x - 2| + |x| = 4$

b) $|x - 2| + |x + 2| = 4$

Let's solve these equations step by step:

a) $|x - 2| + |x| = 4$

We'll consider two cases based on the signs of $x - 2$ and $x$:

Case 1: $x \geq 2$

In this case, $|x - 2| = x - 2$ and $|x| = x$. So the equation becomes:

$x - 2 + x = 4$

Solving for $x$, we get:

$2x - 2 = 4$ $2x = 6$ $x = 3$

Case 2: $x < 2$

In this case, $|x - 2| = -(x - 2)$ and $|x| = -x$. So the equation becomes:

$-(x - 2) - x = 4$

Solving for $x$, we get:

$-x + 2 - x = 4$ $-2x + 2 = 4$ $-2x = 2$ $x = -1$

So, the solutions for equation a) are $x = 3$ and $x = -1$.

b) $|x - 2| + |x + 2| = 4$

We'll again consider two cases based on the signs of $x - 2$ and $x + 2$:

Case 1: $x \geq 2$

In this case, $|x - 2| = x - 2$ and $|x + 2| = x + 2$. So the equation becomes:

$x - 2 + x + 2 = 4$

Solving for $x$, we get:

$2x = 4$ $x = 2$

Case 2: $x < 2$

In this case, $|x - 2| = -(x - 2)$ and $|x + 2| = -(x + 2)$. So the equation becomes:

$-(x - 2) - (x + 2) = 4$

Solving for $x$, we get:

$-x + 2 - x - 2 = 4$ $-2x = 4$ $x = -2$

So, the solutions for equation b) are $x = 2$ and $x = -2$.