Question : What is the value of $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$?
Option 1: $1$
Option 2: $\frac{(x-3)}{(x+3)}$
Option 3: $\frac{(x+4)}{(x-3)}$
Option 4: $\frac{(x-3)}{(x+4)}$
Correct Answer: $1$
Solution : Given: $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$ = $\frac{(x-3)(x+2)}{(x+4)(x-3)}÷\frac{(x+3)(x+2)}{(x+4)(x+3)}$ = $\frac{(x-3)(x+2)}{(x+4)(x-3)}×\frac{(x+4)(x+3)}{(x+3)(x+2)}$ = $1$ Hence, the correct answer is $1$.
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