Answer:
Option (3) is correct.
The given expression [tex]\dfrac{\frac{x}{x-3} }{\frac{x^2}{x^2-9} }[/tex] is equivalent to [tex]frac{x+3}{x}[/tex]
Step-by-step explanation:
Given expression [tex]\dfrac{\frac{x}{x-3} }{\frac{x^2}{x^2-9} }[/tex]
We have to find an equivalent fraction to the given expression [tex]\dfrac{\frac{x}{x-3} }{\frac{x^2}{x^2-9} }[/tex] out of given options.
Consider the given expression [tex]\dfrac{\frac{x}{x-3} }{\frac{x^2}{x^2-9} }[/tex]
Divide fractions [tex]\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}[/tex]
We have,
[tex]=\frac{x\left(x^2-9\right)}{\left(x-3\right)x^2}[/tex]
Cancelling common factor x, we have
[tex]=\frac{\left(x^2-9\right)}{\left(x-3\right)x}[/tex]
Using algebraic identity [tex](a^2-b^2)=(a+b)(a-b)[/tex], we have,
Apply on [tex]x^2-9[/tex] we get, [tex]x^2-3^2=\left(x+3\right)\left(x-3\right)[/tex]
Substitute, we get,
[tex]=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}[/tex]
Cancelling out common factor (x-3) , we get
[tex]=\frac{x+3}{x}[/tex]
Thus, the given expression [tex]\dfrac{\frac{x}{x-3} }{\frac{x^2}{x^2-9} }[/tex] is equivalent to [tex]frac{x+3}{x}[/tex]
Option (3) is correct.
