1.1.3. \( \sqrt{98 x^{6}}+\sqrt{128 x^{6}} \) 1.1.4. \( \frac{y-4}{\sqrt{y}-2} \)

**Problem 1.1.3** We start with: \[ \sqrt{98x^6} + \sqrt{128x^6} \] 1. Factor the radicands: - For \(\sqrt{98x^6}\): \(98 = 49 \cdot 2\) and \(\sqrt{49} = 7\). Also, \(\sqrt{x^6} = x^3\) (assuming \(x \geq 0\)). Thus, \[ \sqrt{98x^6} = \sqrt{49 \cdot 2 \cdot x^6} = \sqrt{49}\sqrt{2x^6} = 7x^3\sqrt{2}. \] 2. Similarly, for \(\sqrt{128x^6}\): - \(128 = 64 \cdot 2\) and \(\sqrt{64} = 8\). Then, \[ \sqrt{128x^6} = \sqrt{64 \cdot 2 \cdot x^6} = 8x^3\sqrt{2}. \] 3. Combine the simplified expressions: \[ 7x^3\sqrt{2} + 8x^3\sqrt{2} = (7 + 8)x^3\sqrt{2} = 15x^3\sqrt{2}. \] **Final answer for 1.1.3:** \[ 15x^3\sqrt{2} \] --- **Problem 1.1.4** We have: \[ \frac{y-4}{\sqrt{y}-2}. \] 1. Recognize that the numerator is a difference of squares since: \[ y-4 = (\sqrt{y})^2 - 2^2. \] Factor the difference of squares: \[ y-4 = (\sqrt{y}-2)(\sqrt{y}+2). \] 2. Substitute into the expression: \[ \frac{(\sqrt{y}-2)(\sqrt{y}+2)}{\sqrt{y}-2}. \] 3. Cancel the common factor \((\sqrt{y}-2)\) (provided that \(\sqrt{y} \neq 2\), i.e., \(y \neq 4\)): \[ \sqrt{y}+2. \] **Final answer for 1.1.4:** \[ \sqrt{y}+2. \]