1. **Problem 2.224 a)**: Add \( \frac{3a^2 + 2}{5a + 5} + \frac{1 - a}{10} \) 2. **Rewrite denominators:** \(5a + 5 = 5(a + 1)\) 3. **Find common denominator:** \(10 = 2 \times 5\), so common denominator is \(10(a + 1)\) 4. **Rewrite each fraction with common denominator:** \[ \frac{3a^2 + 2}{5(a + 1)} = \frac{2(3a^2 + 2)}{10(a + 1)} = \frac{6a^2 + 4}{10(a + 1)} \] \[ \frac{1 - a}{10} = \frac{(1 - a)(a + 1)}{10(a + 1)} \] 5. **Expand numerator of second fraction:** \[ (1 - a)(a + 1) = 1 \cdot a + 1 \cdot 1 - a \cdot a - a \cdot 1 = a + 1 - a^2 - a = 1 - a^2 \] 6. **Sum numerators:** \[ 6a^2 + 4 + 1 - a^2 = 5a^2 + 5 \] 7. **Final fraction:** \[ \frac{5a^2 + 5}{10(a + 1)} = \frac{5(a^2 + 1)}{10(a + 1)} \] 8. **Simplify by canceling 5:** \[ \frac{\cancel{5}(a^2 + 1)}{2 \times \cancel{5}(a + 1)} = \frac{a^2 + 1}{2(a + 1)} \] --- 1. **Problem 2.224 b)**: Add \( \frac{1 + 2b}{8b + 4} - \frac{2b + 9}{8b} \) 2. **Rewrite denominators:** \(8b + 4 = 4(2b + 1)\) 3. **Find common denominator:** \(8b = 8b\), so common denominator is \(8b(2b + 1)\) 4. **Rewrite fractions:** \[ \frac{1 + 2b}{4(2b + 1)} = \frac{2b(1 + 2b)}{8b(2b + 1)} \] \[ \frac{2b + 9}{8b} = \frac{(2b + 9)(2b + 1)}{8b(2b + 1)} \] 5. **Expand numerators:** \[ 2b(1 + 2b) = 2b + 4b^2 \] \[ (2b + 9)(2b + 1) = 4b^2 + 2b + 18b + 9 = 4b^2 + 20b + 9 \] 6. **Subtract numerators:** \[ (2b + 4b^2) - (4b^2 + 20b + 9) = 2b + 4b^2 - 4b^2 - 20b - 9 = -18b - 9 \] 7. **Final fraction:** \[ \frac{-18b - 9}{8b(2b + 1)} = \frac{-9(2b + 1)}{8b(2b + 1)} \] 8. **Cancel \(2b + 1\):** \[ \frac{-9\cancel{(2b + 1)}}{8b\cancel{(2b + 1)}} = \frac{-9}{8b} \] --- 1. **Problem 2.224 c)**: Add \( \frac{12c^2 + 48}{3} + \frac{96 - 12c^3}{3c - 6} \) 2. **Rewrite denominator:** \(3c - 6 = 3(c - 2)\) 3. **Find common denominator:** \(3\) and \(3(c - 2)\) common denominator is \(3(c - 2)\) 4. **Rewrite first fraction:** \[ \frac{12c^2 + 48}{3} = \frac{(12c^2 + 48)(c - 2)}{3(c - 2)} \] 5. **Factor numerator:** \(12c^2 + 48 = 12(c^2 + 4)\) 6. **Expand numerator:** \[ 12(c^2 + 4)(c - 2) = 12(c^3 - 2c^2 + 4c - 8) = 12c^3 - 24c^2 + 48c - 96 \] 7. **Rewrite second fraction numerator:** \(96 - 12c^3 = -12c^3 + 96\) 8. **Sum numerators:** \[ (12c^3 - 24c^2 + 48c - 96) + (-12c^3 + 96) = -24c^2 + 48c \] 9. **Final fraction:** \[ \frac{-24c^2 + 48c}{3(c - 2)} = \frac{24c(-c + 2)}{3(c - 2)} \] 10. **Rewrite numerator:** \(24c(-c + 2) = 24c(2 - c)\) 11. **Note \(2 - c = -(c - 2)\), so:** \[ \frac{24c(2 - c)}{3(c - 2)} = \frac{24c \times -(c - 2)}{3(c - 2)} = \frac{-24c(c - 2)}{3(c - 2)} \] 12. **Cancel \(c - 2\):** \[ \frac{-24c\cancel{(c - 2)}}{3\cancel{(c - 2)}} = \frac{-24c}{3} = -8c \] --- 1. **Problem 2.225 a)**: Add \( \frac{5 - 3g}{8g^2 + 2g} + \frac{5g}{12g^2} \) 2. **Factor denominator:** \(8g^2 + 2g = 2g(4g + 1)\) 3. **Common denominator:** \(12g^2 = 12g^2\), so common denominator is \(12g^2(4g + 1)\) 4. **Rewrite first fraction:** \[ \frac{5 - 3g}{2g(4g + 1)} = \frac{6g(5 - 3g)}{12g^2(4g + 1)} \] 5. **Rewrite second fraction:** \[ \frac{5g}{12g^2} = \frac{5g(4g + 1)}{12g^2(4g + 1)} \] 6. **Expand numerators:** \[ 6g(5 - 3g) = 30g - 18g^2 \] \[ 5g(4g + 1) = 20g^2 + 5g \] 7. **Sum numerators:** \[ (30g - 18g^2) + (20g^2 + 5g) = (30g + 5g) + (-18g^2 + 20g^2) = 35g + 2g^2 \] 8. **Final fraction:** \[ \frac{2g^2 + 35g}{12g^2(4g + 1)} \] --- 1. **Problem 2.225 b)**: Subtract \( \frac{3 - h}{2h + 6} - \frac{3h^2 + 11}{6h^2} \) 2. **Factor denominator:** \(2h + 6 = 2(h + 3)\) 3. **Common denominator:** \(6h^2 = 6h^2\), so common denominator is \(6h^2(h + 3)\) 4. **Rewrite first fraction:** \[ \frac{3 - h}{2(h + 3)} = \frac{3h^2(3 - h)}{6h^2(h + 3)} \] 5. **Rewrite second fraction:** \[ \frac{3h^2 + 11}{6h^2} = \frac{(3h^2 + 11)(h + 3)}{6h^2(h + 3)} \] 6. **Expand numerators:** \[ 3h^2(3 - h) = 9h^2 - 3h^3 \] \[ (3h^2 + 11)(h + 3) = 3h^3 + 9h^2 + 11h + 33 \] 7. **Subtract numerators:** \[ (9h^2 - 3h^3) - (3h^3 + 9h^2 + 11h + 33) = 9h^2 - 3h^3 - 3h^3 - 9h^2 - 11h - 33 = -6h^3 - 11h - 33 \] 8. **Final fraction:** \[ \frac{-6h^3 - 11h - 33}{6h^2(h + 3)} \] --- 1. **Problem 2.225 c)**: Add \( \frac{5e + 6}{15e^2} + \frac{6 - e}{3e^2 - 9e} \) 2. **Factor denominator:** \(3e^2 - 9e = 3e(e - 3)\) 3. **Common denominator:** \(15e^2 = 15e^2\), so common denominator is \(15e^2(e - 3)\) 4. **Rewrite first fraction:** \[ \frac{5e + 6}{15e^2} = \frac{(5e + 6)(e - 3)}{15e^2(e - 3)} \] 5. **Rewrite second fraction:** \[ \frac{6 - e}{3e(e - 3)} = \frac{5e(6 - e)}{15e^2(e - 3)} \] 6. **Expand numerators:** \[ (5e + 6)(e - 3) = 5e^2 - 15e + 6e - 18 = 5e^2 - 9e - 18 \] \[ 5e(6 - e) = 30e - 5e^2 \] 7. **Sum numerators:** \[ (5e^2 - 9e - 18) + (30e - 5e^2) = (5e^2 - 5e^2) + (-9e + 30e) - 18 = 21e - 18 \] 8. **Final fraction:** \[ \frac{21e - 18}{15e^2(e - 3)} = \frac{3(7e - 6)}{15e^2(e - 3)} \] 9. **Simplify by canceling 3:** \[ \frac{\cancel{3}(7e - 6)}{5 \times \cancel{3} e^2 (e - 3)} = \frac{7e - 6}{5 e^2 (e - 3)} \] --- 1. **Problem 2.225 d)**: Subtract \( \frac{2hk - k}{8h^2 - 6h} - \frac{3k}{12h} \) 2. **Factor denominator:** \(8h^2 - 6h = 2h(4h - 3)\) 3. **Common denominator:** \(12h = 12h\), so common denominator is \(12h(4h - 3)\) 4. **Rewrite first fraction:** \[ \frac{2hk - k}{2h(4h - 3)} = \frac{6(2hk - k)}{12h(4h - 3)} \] 5. **Rewrite second fraction:** \[ \frac{3k}{12h} = \frac{3k(4h - 3)}{12h(4h - 3)} \] 6. **Expand numerators:** \[ 6(2hk - k) = 12hk - 6k \] \[ 3k(4h - 3) = 12hk - 9k \] 7. **Subtract numerators:** \[ (12hk - 6k) - (12hk - 9k) = 12hk - 6k - 12hk + 9k = 3k \] 8. **Final fraction:** \[ \frac{3k}{12h(4h - 3)} = \frac{k}{4h(4h - 3)} \] --- 1. **Problem 2.225 e)**: Subtract \( \frac{2hk^3 + 5k}{7k^2} - \frac{8hk + 11k}{14 - 7k^2} \) 2. **Factor denominator:** \(14 - 7k^2 = 7(2 - k^2)\) 3. **Common denominator:** \(7k^2\) and \(7(2 - k^2)\) common denominator is \(7k^2(2 - k^2)\) 4. **Rewrite first fraction:** \[ \frac{2hk^3 + 5k}{7k^2} = \frac{(2hk^3 + 5k)(2 - k^2)}{7k^2(2 - k^2)} \] 5. **Rewrite second fraction:** \[ \frac{8hk + 11k}{7(2 - k^2)} = \frac{k^2(8h + 11)}{7k^2(2 - k^2)} \] 6. **Expand numerator of first fraction:** \[ (2hk^3 + 5k)(2 - k^2) = 4hk^3 - 2hk^5 + 10k - 5k^3 \] 7. **Numerator of second fraction:** \[ k^2(8h + 11) = 8hk^2 + 11k^2 \] 8. **Subtract numerators:** \[ (4hk^3 - 2hk^5 + 10k - 5k^3) - (8hk^2 + 11k^2) = -2hk^5 + 4hk^3 - 8hk^2 + 10k - 5k^3 - 11k^2 \] 9. **Final fraction:** \[ \frac{-2hk^5 + 4hk^3 - 8hk^2 + 10k - 5k^3 - 11k^2}{7k^2(2 - k^2)} \] --- 1. **Problem 2.225 f)**: Add \( \frac{1}{15} + \frac{4k^2 - 3}{3k^2 - 9kg} \) 2. **Factor denominator:** \(3k^2 - 9kg = 3k(k - 3g)\) 3. **Common denominator:** \(15 = 3 \times 5\), so common denominator is \(15k(k - 3g)\) 4. **Rewrite first fraction:** \[ \frac{1}{15} = \frac{k(k - 3g)}{15k(k - 3g)} \] 5. **Rewrite second fraction:** \[ \frac{4k^2 - 3}{3k(k - 3g)} = \frac{5(4k^2 - 3)}{15k(k - 3g)} \] 6. **Sum numerators:** \[ k(k - 3g) + 5(4k^2 - 3) = k^2 - 3kg + 20k^2 - 15 = 21k^2 - 3kg - 15 \] 7. **Final fraction:** \[ \frac{21k^2 - 3kg - 15}{15k(k - 3g)} \] --- **Slug:** "bruchterme addieren" **Subject:** "algebra" **svg:** "" **desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 10