) 13sqrt(7) + 10sqrt(28) - 20sqrt(1, 75) 5) (x ^ 2 + 1 - x * sqrt(3))(x ^ 2 + 1 + x * sqrt(3)) B) (5 - sqrt(3)) ^ 2 + (2 - 3sqrt(3)) ^ 2 - (2sqrt(3) - 1) ^ 2 г 1sqrt(28 - 10sqrt(3)) + sqrt(31 - 12sqrt(3)) - sqrt(13 - 4sqrt(3))

Let's solve each problem step by step. ### 1) Simplify \( 13\sqrt{7} + 10\sqrt{28} - 20\sqrt{1.75} \) First, simplify the square roots: - \( \sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7} \) - \( \sqrt{1.75} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \) Now substitute these back into the expression: \[ 13\sqrt{7} + 10(2\sqrt{7}) - 20\left(\frac{\sqrt{7}}{2}\right) \] This simplifies to: \[ 13\sqrt{7} + 20\sqrt{7} - 10\sqrt{7} \] Combine the terms: \[ (13 + 20 - 10)\sqrt{7} = 23\sqrt{7} \] ### Answer: \( 23\sqrt{7} \) --- ### 5) Expand \((x^2 + 1 - x\sqrt{3})(x^2 + 1 + x\sqrt{3})\) This expression is a difference of squares: \[ (a - b)(a + b) = a^2 - b^2 \text{ where } a = x^2 + 1 \text{ and } b = x\sqrt{3} \] Calculating \( a^2 \) and \( b^2 \): \[ a^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1 \] \[ b^2 = (x\sqrt{3})^2 = 3x^2 \] Putting it into the difference of squares formula: \[ (x^2 + 1)^2 - (x\sqrt{3})^2 = (x^4 + 2x^2 + 1) - 3x^2 = x^4 - x^2 + 1 \] ### Answer: \( x^4 - x^2 + 1 \) --- ### B) Simplify \((5 - \sqrt{3})^2 + (2 - 3\sqrt{3})^2 - (2\sqrt{3} - 1)^2\) Calculate each square: 1. \((5 - \sqrt{3})^2 = 25 - 10\sqrt{3} + 3 = 28 - 10\sqrt{3}\) 2. \((2 - 3\sqrt{3})^2 = 4 - 12\sqrt{3} + 27 = 31 - 12\sqrt{3}\) 3. \((2\sqrt{3} - 1)^2 = 12 - 4\sqrt{3} + 1 = 13 - 4\sqrt{3}\) Now put it all together: \[ (28 - 10\sqrt{3}) + (31 - 12\sqrt{3}) - (13 - 4\sqrt{3}) \] Combine like terms: \[ (28 + 31 - 13) + (-10\sqrt{3} - 12\sqrt{3} + 4\sqrt{3}) = 46 - 18\sqrt{3} \] ### Answer: \( 46 - 18\sqrt{3} \) --- ### г) Simplify \( \sqrt{28 - 10\sqrt{3}} + \sqrt{31 - 12\sqrt{3}} - \sqrt{13 - 4\sqrt{3}} \) For each square root, we need to simplify: 1. \( \sqrt{28 - 10\sqrt{3}} \) - This can be expressed as \( \sqrt{(a - b)^2} \) where \( a^2 + b^2 = 28 \) and \( 2ab = 10\sqrt{3} \). - Solve \( ab = 5\sqrt{3} \). - Let \( a = 5 \) and \( b = \sqrt{3} \) results in: \[ (5^2 + (\sqrt{3})^2 = 25 + 3 = 28) \rightarrow \sqrt{28 - 10\sqrt{3}} = 5 - \sqrt{3} \] 2. \( \sqrt{31 - 12\sqrt{3}} \): - Similar structure gives \(a = 6\) and \( b = \sqrt{3}\) - Result: \(6 - \sqrt{3}\). 3. \( \sqrt{13 - 4\sqrt{3}} \): - Gives \(a = 4\) and \(b = \sqrt{3}\) - Result: \(4 - \sqrt{3}\). Combine: \[ (5-\sqrt{3}) + (6-\sqrt{3}) - (4-\sqrt{3}) = 5 + 6 - 4 - \sqrt{3} - \sqrt{3} + \sqrt{3} = 7 - \sqrt{3} \] ### Answer: \( 7 - \sqrt{3} \) --- Feel free to ask more questions or request clarifications on any part!