|4х+1|=0 |х:2-1|+|х2-4|=0 |х-5|=|2х+3|

Let's solve each equation step by step.

1. Equation: \(|4x + 1| = 0\)

   The absolute value of any expression is zero only if the expression itself is zero. Therefore, we solve: \[ 4x + 1 = 0 \] Solving for \(x\): \[ 4x = -1 \] \[ x = -\frac{1}{4} \] Solution: \(x = -\frac{1}{4}\)

2. Equation: \(|\frac{x}{2} - 1| + |x^2 - 4| = 0\)

   The sum of absolute values is zero only if both absolute values are zero. Therefore, we solve: \[ \frac{x}{2} - 1 = 0 \quad \text{and} \quad x^2 - 4 = 0 \]

   From the first equation: \[ \frac{x}{2} = 1 \Rightarrow x = 2 \]

   From the second equation: \[ x^2 - 4 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2 \text{ or } x = -2 \]

   For both conditions to hold simultaneously, we need to check if both solutions work:

   • For \(x = 2\): \(|\frac{2}{2} - 1| + |2^2 - 4| = |1 - 1| + |4 - 4| = |0| + |0| = 0\), which satisfies.
   • For \(x = -2\): \(|\frac{-2}{2} - 1| + |-2^2 - 4| = | -1 - 1| + |4 - 4| = | -2| + |0| = 2 \neq 0\), which does not satisfy.

   Solution: \(x = 2\)

3. Equation: \(|x - 5| = |2x + 3|\)

   We can break this into two cases based on the nature of absolute values:

   Case 1: \(x - 5 = 2x + 3\) \[ x - 2x = 3 + 5 \] \[ -x = 8 \Rightarrow x = -8 \]

   Case 2: \(x - 5 = -(2x + 3)\) \[ x - 5 = -2x - 3 \] \[ x + 2x = 5 - 3 \] \[ 3x = 2 \Rightarrow x = \frac{2}{3} \]

   Now, we test these solutions:

   • For \(x = -8\): \(|-8 - 5| = |2(-8) + 3| \Rightarrow | -13| = |-16 + 3| \Rightarrow 13 = 13\) (true)
   • For \(x = \frac{2}{3}\): \(|\frac{2}{3} - 5| = |2(\frac{2}{3}) + 3| \Rightarrow |-\frac{13}{3}| = |\frac{4}{3} + 3| \Rightarrow \frac{13}{3} = |\frac{13}{3}| \) (true)

   Solutions: \(x = -8\) and \(x = \frac{2}{3}\)

To summarize:

1. \(x = -\frac{1}{4}\)
2. \(x = 2\)
3. \(x = -8\) and \(x = \frac{2}{3}\)