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15. $\frac{x^{2}+x - 12}{x^{2}+x - 30}cdot\frac{x^{2}+5x + 6}{x^{2}-2x …

Question

  1. $\frac{x^{2}+x - 12}{x^{2}+x - 30}cdot\frac{x^{2}+5x + 6}{x^{2}-2x - 3}div\frac{x + 3}{x^{2}+7x + 6}$
  2. $\frac{\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}{\frac{4a - 4b}{2a^{2}+2b^{2}}}$
  3. $\frac{\frac{p^{3}-8q^{3}}{p^{2}-4q^{2}}}{\frac{p^{2}+4pq + 4q^{2}}{(p + 2q)^{2}}}$
  4. $\frac{3x^{2}y^{4}}{4}cdot\frac{2}{9x^{3}y}$
  5. $\frac{5ab^{3}}{3b^{4}}div\frac{10a^{2}b^{8}}{6b^{2}}$
  6. $\frac{x + 1}{3}div\frac{3x + 3}{7}$

Explanation:

Response
  1. For problem 15:
  • Explanation:
  • Step 1: Factor the quadratic - expressions
  • \(x^{2}+x - 12=(x + 4)(x-3)\), \(x^{2}+x - 30=(x + 6)(x - 5)\), \(x^{2}+5x + 6=(x + 2)(x+3)\), \(x^{2}-2x - 3=(x - 3)(x + 1)\), \(x^{2}+7x + 6=(x + 1)(x + 6)\).
  • The original expression \(\frac{x^{2}+x - 12}{x^{2}+x - 30}\cdot\frac{x^{2}+5x + 6}{x^{2}-2x - 3}\div\frac{x + 3}{x^{2}+7x + 6}\) becomes \(\frac{(x + 4)(x - 3)}{(x + 6)(x - 5)}\cdot\frac{(x + 2)(x + 3)}{(x - 3)(x + 1)}\cdot\frac{(x + 1)(x + 6)}{x + 3}\).
  • Step 2: Cancel out the common factors
  • Cancel out \((x - 3)\), \((x + 3)\), \((x + 1)\) and \((x + 6)\) in the numerator and denominator.
  • The result is \(\frac{(x + 4)(x + 2)}{x - 5}=\frac{x^{2}+6x + 8}{x - 5}\).
  1. For problem 16:
  • Explanation:
  • Step 1: Simplify the complex - fraction
  • First, rewrite the complex - fraction \(\frac{\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}{\frac{4a - 4b}{2a^{2}+2b^{2}}}\) as \(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\cdot\frac{2a^{2}+2b^{2}}{4a - 4b}\).
  • Factor \(a^{2}-b^{2}=(a + b)(a - b)\), \(2a^{2}+2b^{2}=2(a^{2}+b^{2})\), and \(4a - 4b = 4(a - b)\).
  • The expression becomes \(\frac{(a + b)(a - b)}{a^{2}+b^{2}}\cdot\frac{2(a^{2}+b^{2})}{4(a - b)}\).
  • Step 2: Cancel out the common factors
  • Cancel out \((a - b)\) and \((a^{2}+b^{2})\).
  • \(\frac{(a + b)\cdot2}{4}=\frac{a + b}{2}\).
  1. For problem 17:
  • Explanation:
  • Step 1: Factor the expressions
  • Recall \(p^{3}-8q^{3}=(p - 2q)(p^{2}+2pq + 4q^{2})\), \(p^{2}-4q^{2}=(p + 2q)(p - 2q)\), \(p^{2}+4pq + 4q^{2}=(p + 2q)^{2}\).
  • The original expression \(\frac{\frac{p^{3}-8q^{3}}{p^{2}-4q^{2}}}{\frac{p^{2}+4pq + 4q^{2}}{(p + 2q)^{2}}}\) is \(\frac{p^{3}-8q^{3}}{p^{2}-4q^{2}}\cdot\frac{(p + 2q)^{2}}{p^{2}+4pq + 4q^{2}}\).
  • Substitute the factored forms: \(\frac{(p - 2q)(p^{2}+2pq + 4q^{2})}{(p + 2q)(p - 2q)}\cdot\frac{(p + 2q)^{2}}{(p + 2q)^{2}}\).
  • Step 2: Cancel out the common factors
  • Cancel out \((p - 2q)\) and \((p + 2q)^{2}\).
  • The result is \(\frac{p^{2}+2pq + 4q^{2}}{p + 2q}\).
  1. For problem 18:
  • Explanation:
  • Step 1: Multiply the fractions
  • Multiply \(\frac{3x^{2}y^{4}}{4}\cdot\frac{2}{9x^{3}y}\).
  • Multiply the numerators together: \(3x^{2}y^{4}\cdot2 = 6x^{2}y^{4}\), and the denominators together: \(4\cdot9x^{3}y=36x^{3}y\).
  • The expression is \(\frac{6x^{2}y^{4}}{36x^{3}y}\).
  • Step 2: Simplify the fraction using the rules of exponents
  • Use the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\).
  • \(\frac{6}{36}=\frac{1}{6}\), \(\frac{x^{2}}{x^{3}}=\frac{1}{x}\), \(\frac{y^{4}}{y}=y^{3}\).
  • The result is \(\frac{y^{3}}{6x}\).
  1. For problem 19:
  • Explanation:
  • Step 1: Rewrite the division as multiplication by the reciprocal
  • \(\frac{5ab^{3}}{3b^{4}}\div\frac{10a^{2}b^{8}}{6b^{2}}\) is \(\frac{5ab^{3}}{3b^{4}}\cdot\frac{6b^{2}}{10a^{2}b^{8}}\).
  • Step 2: Multiply the numerators and denominators
  • Numerator: \(5ab^{3}\cdot6b^{2}=30ab^{5}\).
  • Denominator: \(3b^{4}\cdot10a^{2}b^{8}=30a^{2}b^{12}\).
  • The expression is \(\frac{30ab^{5}}{30a^{2}b^{12}}\).
  • Step 3: Simplify the fraction
  • \(\frac{30}{30}=1\), \(\frac{a}{a^{2}}=\frac{1}{a}\), \(\frac{b^{5}}{b^{12}}=\frac{1}{b^{7}}\).
  • The result is \(\frac{1}{ab^{7}}\).
  1. For problem 20:
  • Explanation:
  • **Step 1: Rewrite the division as multiplication by the recipro…

Answer:

  1. \(\frac{x^{2}+6x + 8}{x - 5}\)
  2. \(\frac{a + b}{2}\)
  3. \(\frac{p^{2}+2pq + 4q^{2}}{p + 2q}\)
  4. \(\frac{y^{3}}{6x}\)
  5. \(\frac{1}{ab^{7}}\)
  6. \(\frac{7}{9}\)