Divide Rational Expressions
Simplify the rational expressions:
- $\require{cancel}\dfrac{x}{x^2}\div \dfrac{x}{x^2}=\dfrac{\cancel{x}}{\cancel{x^2}}\div \dfrac{\cancel{x^2}}{\cancel{x}}=\mathbf{1}$
- $\dfrac{x}{3x}\div \dfrac{x}{2x}= \dfrac{\cancel{x}}{3\cancel{x}}\div \dfrac{2\cancel{x}}{\cancel{x}}=\mathbf{\dfrac{2}{3}}$
- $\dfrac{x+1}{x^2}\div \dfrac{x+1}{7x}=\dfrac{\cancel{x+1}}{\underset{x}{\cancel{x^2}}}\div \dfrac{7\cancel{x}}{\cancel{x+1}}=\mathbf{\dfrac{7}{x}}$
- $\dfrac{9x}{3x^2+6x-9}\div \dfrac{x^2}{3x^2-27}=\dfrac{9\cancel{x}}{\cancel{3}\cancel{(x+3)}(x-1)}\div \dfrac{\cancel{3}\cancel{(x+3)}(x-3)}{\underset{x}{\cancel{x^2}}}=\mathbf{\dfrac{9(x-3)}{x(x-1)}}$
- $\dfrac{2x-2}{7x^2+14x}\div \dfrac{2x}{x+2}=\dfrac{\cancel{2}(x-1)}{7x\cancel{(x+2)}}\div \dfrac{\cancel{x+2}}{\cancel{2}x}=\mathbf{\dfrac{x-1}{7x^2}}$
- $\dfrac{x^2-5x-24}{x^2+2x+1}\div \dfrac{x^2-8x}{x+1}=\dfrac{\cancel{(x-8)}(x+3)}{(x+1)\cancel{(x+1)}}\div \dfrac{\cancel{x+1}}{x\cancel{(x-8)}}=\mathbf{\dfrac{x+3}{x(x+1)}}$
- $\dfrac{9x}{x^2+4x-21}\div \dfrac{3x^2+9x}{x^2}=\dfrac{\overset{3}{\cancel{9}\cancel{x}}}{(x-3)(x+7)}\div \dfrac{x^2}{\cancel{3x}(x+3)}=\mathbf{\dfrac{3x^2}{(x^2-9)(x+7)}}$
- $\dfrac{x^3-3x^2-10x}{x^2+5x-36}\div \dfrac{10x}{x^2+7x-18}=\dfrac{\cancel{x}(x+2)(x-5)}{(x-4)\cancel{(x+9)}}\div \dfrac{(x-2)\cancel{(x+9)}}{10\cancel{x}}=\mathbf{\dfrac{(x^2-4)(x-5)}{10(x-4)}}$