Sinplify a) \( \frac{a b-a^{2}}{b^{2}-a} \div \frac{a^{2}}{b^{2}+a b} \) b. b. \( \frac{a^{2}-a}{a}-\frac{a}{18} \) c. \( \frac{a-4}{2(a-a)} \) d) \( \frac{x^{2}+2 x+1}{x^{2}+x} \)
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To simplify the expressions, let’s dive right in! For a) To simplify \( \frac{a b-a^{2}}{b^{2}-a} \div \frac{a^{2}}{b^{2}+a b} \), first, let's rewrite the division as multiplication by the reciprocal: \[ \frac{a b-a^{2}}{b^{2}-a} \cdot \frac{b^{2}+a b}{a^{2}} \] Now, factor \( a b-a^{2} \) as \( a(b-a) \) and rewrite \( b^{2}-a \) as \( (b-\sqrt{a})(b+\sqrt{a}) \). This leads us to: \[ \frac{a(b-a)(b^{2}+ab)}{(b^{2}-a)a^{2}} \] After canceling common factors, the final expression simplifies accordingly. For b) In the expression \( \frac{a^{2}-a}{a}-\frac{a}{18} \), factor the first term: \[ \frac{a(a-1)}{a} - \frac{a}{18} = a - \frac{a}{18} \] Finding a common denominator of 18 gives: \[ \frac{18a - a}{18} = \frac{17a}{18} \] This is the simplified form! For c) When you look at \( \frac{a-4}{2(a-a)} \), the denominator becomes \( 2 \cdot 0 \), which means it’s undefined since division by zero is a no-go in math. For d) For the last expression \( \frac{x^{2}+2 x+1}{x^{2}+x} \), notice that \( x^{2}+2x+1 \) can be factored as \( (x+1)^{2} \) while the denominator \( x^{2}+x \) simplifies to \( x(x+1) \). This leads to: \[ \frac{(x+1)^{2}}{x(x+1)} = \frac{x+1}{x} \] as long as \( x \neq -1 \) (because we can't divide by zero). There you have it—a simplified path through each of these expressions!
