Evaluate √147 - √108 + √75 / √243 - √192

Use the identity $(a+b)(a-b) = a^2 - b^2$ for the denominator.

$$\text{Numerator} = (3\sqrt{3} + 2\sqrt{2})(2\sqrt{3} - 3\sqrt{2})$$ $$= 3\sqrt{3} \cdot 2\sqrt{3} - 3\sqrt{3} \cdot 3\sqrt{2} + 2\sqrt{2} \cdot 2\sqrt{3} - 2\sqrt{2} \cdot 3\sqrt{2}$$ $$= 6(3) - 9\sqrt{6} + 4\sqrt{6} - 6(2)$$ $$= 18 - 9\sqrt{6} + 4\sqrt{6} - 12$$ $$= 6 - 5\sqrt{6}$$ $$\text{Denominator} = (2\sqrt{3})^2 - (3\sqrt{2})^2$$ $$= 4(3) - 9(2)$$ $$= 12 - 18$$ $$= -6$$

$$\frac{6 - 5\sqrt{6}}{-6} = -1 + \frac{5\sqrt{6}}{6}$$

The final answer is $\boxed{-1 + \frac{5\sqrt{6}}{6}}$.

:
Given $x * 3 = \frac{1}{2}$, find the value of $x$. The operation is $p * q = \frac{3p+2q}{2p+3q}$.
Here, $p = x$ and $q = 3$.

$$\frac{3x + 2(3)}{2x + 3(3)} = \frac{1}{2}$$ $$\frac{3x + 6}{2x + 9} = \frac{1}{2}$$

$$2(3x + 6) = 1(2x + 9)$$ $$6x + 12 = 2x + 9$$

$$6x - 2x = 9 - 12$$ $$4x = -3$$

$$x = -\frac{3}{4}$$

The final answer is $\boxed{-\frac{3}{4}}$.

:
Simplify $\sqrt{3125} - 4\sqrt{128} + 5\sqrt{80}$.

$$\sqrt{3125} = \sqrt{625 \times 5} = 25\sqrt{5}$$ $$\sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$$ $$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$$

$$25\sqrt{5} - 4(8\sqrt{2}) + 5(4\sqrt{5})$$ $$25\sqrt{5} - 32\sqrt{2} + 20\sqrt{5}$$

$$(25\sqrt{5} + 20\sqrt{5}) - 32\sqrt{2}$$ $$45\sqrt{5} - 32\sqrt{2}$$

The final answer is $\boxed{45\sqrt{5} - 32\sqrt{2}}$.

:
Simplify $\frac{4\sqrt{2} + 3\sqrt{5}}{\sqrt{10}}$.

$$\frac{4\sqrt{2}\sqrt{10} + 3\sqrt{5}\sqrt{10}}{10}$$ $$\frac{4\sqrt{20} + 3\sqrt{50}}{10}$$

$$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$$ $$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$

$$\frac{4(2\sqrt{5}) + 3(5\sqrt{2})}{10}$$ $$\frac{8\sqrt{5} + 15\sqrt{2}}{10}$$

The final answer is $\boxed{\frac{8\sqrt{5} + 15\sqrt{2}}{10}}$.

:
Simplify $\frac{\sqrt{147} - \sqrt{108} + \sqrt{75}}{\sqrt{243} - \sqrt{192}}$.

$$\sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}$$ $$\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$$ $$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$$

$$\sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$$ $$\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$$

$$\frac{(7 - 6 + 5)\sqrt{3}}{(9 - 8)\sqrt{3}}$$ $$\frac{6\sqrt{3}}{1\sqrt{3}}$$

$$\frac{6}{1} = 6$$

The final answer is $\boxed{6}$.

:
Simplify $\frac{\sqrt{8} + 5\sqrt{32} - 4\sqrt{50}}{1 - \sqrt{18}}$.

$$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$$ $$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$$ $$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$

$$\frac{2\sqrt{2} + 5(4\sqrt{2}) - 4(5\sqrt{2})}{1 - 3\sqrt{2}}$$ $$\frac{2\sqrt{2} + 20\sqrt{2} - 20\sqrt{2}}{1 - 3\sqrt{2}}$$

$$\frac{(2 + 20 - 20)\sqrt{2}}{1 - 3\sqrt{2}}$$ $$\frac{2\sqrt{2}}{1 - 3\sqrt{2}}$$

Use the identity $(a-b)(a+b) = a^2 - b^2$ for the denominator.

$$\text{Numerator} = 2\sqrt{2}(1 + 3\sqrt{2})$$ $$= 2\sqrt{2} \cdot 1 + 2\sqrt{2} \cdot 3\sqrt{2}$$ $$= 2\sqrt{2} + 6(2)$$ $$= 2\sqrt{2} + 12$$ $$\text{Denominator} = (1)^2 - (3\sqrt{2})^2$$ $$= 1 - 9(2)$$ $$= 1 - 18$$ $$= -17$$

$$\frac{12 + 2\sqrt{2}}{-17} = -\frac{12 + 2\sqrt{2}}{17}$$

The final answer is $\boxed{-\frac{12 + 2\sqrt{2}}{17}}$.