2у²-16=0 5х-2х²=0 -4х+16=0 (У-1) ²+2у=3(у+1/3)

Let's solve each of the equations step by step.

1. Equation: \(2y^2 - 16 = 0\)

   To solve for \(y\): \[ 2y^2 = 16 \] \[ y^2 = 8 \] \[ y = \pm \sqrt{8} = \pm 2\sqrt{2} \]

   Solutions: \(y = 2\sqrt{2}, -2\sqrt{2}\)

2. Equation: \(5x - 2x^2 = 0\)

   We can factor this equation: \[ x(5 - 2x) = 0 \] Setting each factor to zero gives: \[ x = 0 \quad \text{or} \quad 5 - 2x = 0 \] Solving \(5 - 2x = 0\): \[ 2x = 5 \quad \Rightarrow \quad x = \frac{5}{2} \]

   Solutions: \(x = 0, \frac{5}{2}\)

3. Equation: \(-4x + 16 = 0\)

   To solve for \(x\): \[ -4x = -16 \] \[ x = 4 \]

   Solution: \(x = 4\)

4. Equation: \((y - 1)^2 + 2y = 3\left(y + \frac{1}{3}\right)\)

   First, simplify the right side: \[ \Rightarrow 3y + 1 \] Now we rewrite the equation: \[ (y - 1)^2 + 2y = 3y + 1 \]

   Expanding the left side: \[ y^2 - 2y + 1 + 2y = y^2 + 1 \] This leads to: \[ y^2 + 1 = 3y + 1 \] Subtract \(1\) from both sides: \[ y^2 = 3y \] Rearranging this gives: \[ y^2 - 3y = 0 \] Factoring: \[ y(y - 3) = 0 \] Setting each factor to zero gives: \[ y = 0 \quad \text{or} \quad y = 3 \]

   Solutions: \(y = 0, 3\)

Summary of solutions:

1. \(y = 2\sqrt{2}, -2\sqrt{2}\)
2. \(x = 0, \frac{5}{2}\)
3. \(x = 4\)
4. \(y = 0, 3\)