$$ 1.19)(\sqrt{5}+4)^{2} $$ 1.14) $$ \frac{2 \sqrt{12}-\sqrt{48}}{\sqrt{3}} $$ 1.15) $$ \frac{\sqrt[3]{16}+\sqrt[3]{54}}{\sqrt[3]{2}} $$ $$ 1.023) \sqrt[3]{6}(\sqrt[3]{4}-\sqrt[3]{9}) $$

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Let's simplify and evaluate each of the given expressions step by step.

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### 1.19) $$ 1.19 (\sqrt{5} + 4)^2 $$

**Step 1: Expand the Binomial**

$$
(\sqrt{5} + 4)^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot 4 + 4^2 = 5 + 8\sqrt{5} + 16 = 21 + 8\sqrt{5}
$$

**Step 2: Multiply by 1.19**

$$
1.19 \times (21 + 8\sqrt{5}) = 1.19 \times 21 + 1.19 \times 8\sqrt{5} = 24.99 + 9.52\sqrt{5}
$$

**Numerical Approximation:**

$$
\sqrt{5} \approx 2.236 \
9.52 \times 2.236 \approx 21.3 \
24.99 + 21.3 \approx 46.29
$$

**Final Answer:**
$$
1.19 (\sqrt{5} + 4)^2 = 24.99 + 9.52\sqrt{5} \approx 46.29
$$

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### 1.14) $$ \frac{2 \sqrt{12} - \sqrt{48}}{\sqrt{3}} $$

**Step 1: Simplify the Square Roots**

$$
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \
\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}
$$

**Step 2: Substitute Back into the Expression**

$$
\frac{2 \times 2\sqrt{3} - 4\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3} - 4\sqrt{3}}{\sqrt{3}} = \frac{0}{\sqrt{3}} = 0
$$

**Final Answer:**
$$
\frac{2 \sqrt{12} - \sqrt{48}}{\sqrt{3}} = 0
$$

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### 1.15) $$ \frac{\sqrt[3]{16} + \sqrt[3]{54}}{\sqrt[3]{2}} $$

**Step 1: Simplify the Cube Roots**

$$
\sqrt[3]{16} = \sqrt[3]{8 \times 2} = 2\sqrt[3]{2} \
\sqrt[3]{54} = \sqrt[3]{27 \times 2} = 3\sqrt[3]{2}
$$

**Step 2: Substitute Back into the Expression**

$$
\frac{2\sqrt[3]{2} + 3\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{5\sqrt[3]{2}}{\sqrt[3]{2}} = 5
$$

**Final Answer:**
$$
\frac{\sqrt[3]{16} + \sqrt[3]{54}}{\sqrt[3]{2}} = 5
$$

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### 1.023) $$ \sqrt[3]{6} \left( \sqrt[3]{4} - \sqrt[3]{9} \right) $$

**Step 1: Simplify Each Cube Root**

$$
\sqrt[3]{6} = \sqrt[3]{2 \times 3} = \sqrt[3]{2} \cdot \sqrt[3]{3} \
\sqrt[3]{4} = \sqrt[3]{2^2} = (\sqrt[3]{2})^2 \
\sqrt[3]{9} = \sqrt[3]{3^2} = (\sqrt[3]{3})^2
$$

**Step 2: Substitute and Expand**

Let $$ x = \sqrt[3]{2} $$ and $$ y = \sqrt[3]{3} $$, then:

$$
\sqrt[3]{6} (\sqrt[3]{4} - \sqrt[3]{9}) = xy (x^2 - y^2) = xyx^2 - xy y^2 = x^3 y - x y^3
$$

Since $$ x^3 = 2 $$ and $$ y^3 = 3 $$:

$$
2y - 3x = 2\sqrt[3]{3} - 3\sqrt[3]{2}
$$

**Final Answer:**
$$
\sqrt[3]{6} \left( \sqrt[3]{4} - \sqrt[3]{9} \right) = 2\sqrt[3]{3} - 3\sqrt[3]{2}
$$

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1.19) $$ 1.19 (\sqrt{5} + 4)^2 \approx 46.29 $$ 1.14) $$ \frac{2 \sqrt{12} - \sqrt{48}}{\sqrt{3}} = 0 $$ 1.15) $$ \frac{\sqrt[3]{16} + \sqrt[3]{54}}{\sqrt[3]{2}} = 5 $$ 1.023) $$ \sqrt[3]{6} (\sqrt[3]{4} - \sqrt[3]{9}) = 2\sqrt[3]{3} - 3\sqrt[3]{2} $$

\( 1.19)(\sqrt{5}+4)^{2} \) 1.14) \( \frac{2 \sqrt{12}-\sqrt{48}}{\sqrt{3}} \) 1.15) \( \frac{\sqrt[3]{16}+\sqrt[3]{54}}{\sqrt[3]{2}} \) \( 1.023) \sqrt[3]{6}(\sqrt[3]{4}-\sqrt[3]{9}) \)

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