Question

1) 7|x| - 6 |x| = 1; 3) |-x| + 4-x| = 9; 2) 21x| + 3|x| = 8; 4) 5-х — 2-x| = 18;— 6)5-- 7, ​

Answer 1

Ответ:

если я правильно понял

1) 7|x|-6|x|=1

x=1

2)21|x|+3|x|=8

24|x|=8

|x|=24:8

|x|=3 или |x|=-3

3)|-x|+4|-x|=9

|-2x|+4=9

|-2x|=9-4

|-2x|=5

|-x|=5:2

|-x|=2,5 или |х|=-2,5.

4)5|-х|-2|-х|=18

|-2х|+5-2=18

|-2х|+5=18+2

|-2х|+5=20

|-2х|=20-5

|-2х|=15

|-х|=15:2

|-х|=7,5 или |х|=-7,5

Пошаговое объяснение:

надеюсь я правильно понял задание, удачи✨

Answer 2

It looks like you've provided a list of equations involving absolute values. You seem to want to solve these equations for the variable "x". Let's go through each equation one by one:

1. $7|x| - 6|x| = 1$ Combine the absolute value terms on the left side: $7|x| - 6|x| = |x|$. So the equation becomes: $|x| = 1$

This means that $x$ could be either 1 or -1.

2. $21x| + 3|x| = 8$ Combine the absolute value terms on the left side: $21x| + 3|x| = (21x + 3)|x|$. So the equation becomes: $(21x + 3)|x| = 8$

To solve this equation, you need to consider cases where $x$ is positive or negative. If $x$ is positive, then $|x| = x$, and if $x$ is negative, then $|x| = -x$. So we have two cases:

Case 1: $x > 0$ $21x + 3 = 8$ Solve for $x$: $21x = 5$ → $x = \frac{5}{21}$

Case 2: $x < 0$ $21x - 3 = 8$ Solve for $x$: $21x = 11$ → $x = \frac{11}{21}$

So, the solutions are $x = \frac{5}{21}$ and $x = \frac{11}{21}$.

3. $|-x| + 4 - |x| = 9$ Combine the constants: $4 - 9 = -5$, so the equation becomes: $|-x| - |x| = -5$

This equation doesn't have a solution for real values of $x$, as the absolute value of a number is always non-negative, and the sum of two non-negative numbers can't be negative.

4. $5 - x - 2 - x| = 18$ Combine the constants: $5 - 2 = 3$, so the equation becomes: $3 - x - x| = 18$

This equation also doesn't have a solution for real values of $x$, as the absolute value of $x$ can't make the left side reach 18.

5. $5 - (-7)$ This is a simple subtraction: $5 - (-7) = 12$.

If you have more equations or if you need further clarification, feel free to ask!