Verified Answer :

One possible equation is:

\(\frac{3(6-(4\div 2)^2)+9-2^3}{11-(2\times 5)}\)

To double check that this is correct, work from the inside of the numerator out:

\(\frac{3(6-(4\div2)^2)+9-2^3}{11-(2\times5)}
\\
\\=\frac{3(6-(2)^2)+9-2^3}{11-(2\times5)}
\\
\\=\frac{3(6-4)+9-2^3}{11-(2\times5)}
\\
\\=\frac{3(2)+9-2^3}{11-(2\times5)}
\\
\\=\frac{3(2)+9-8}{11-(2\times5)}
\\
\\=\frac{6+9-8}{11-(2\times5)}
\\
\\=\frac{7}{11-(10)}
\\
\\=\frac{7}{1}=7\)

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