Verified Answer :

nobillionaireNobley

\(\frac{\frac{1}{9}-\frac{1}{x}}{\frac{1}{81}-\frac{1}{x^2}}=\frac{\frac{x-9}{9x}}{\frac{x^2-81}{81x^2}}=\frac{81x^2(x-9)}{9x(x-9)(x+9)}=\frac{9x}{x+9}\)

Answer:

\(\dfrac{9x}{x+9}\)

Step-by-step explanation:

we have to simplify:

 \(\dfrac{\dfrac{1}{9}-\dfrac{1}{x}}{\dfrac{1}{81}-\dfrac{1}{x^2}}\\ \\\\=\dfrac{\dfrac{x-9}{9x}}{\dfrac{x^2-81}{81x^2}}\\ \\\\=9x\times \dfrac{x-9}{x^2-81}\\ \\\\=9x\times \dfrac{x-9}{(x-9)(x+9)}\\ \\\\=\dfrac{9x}{x+9}\)

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