To solve the exercise you can use the following properties of exponents and radicals
[tex]a^{\frac{m}{n}}=\sqrt[n]{a^m}[/tex][tex]a^ma^n=a^{m+n}[/tex][tex]\frac{a^m}{a^n}=a^{m-n}[/tex]
So, you have
[tex]\begin{gathered} \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=\frac{7^{\frac{1}{3}}\cdot7^{\frac{1}{2}}}{\sqrt[6]{7^5}} \\ \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=\frac{7^{\frac{1}{3}+\frac{1}{2}}}{7^{\frac{5}{6}}} \\ \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=\frac{7^{\frac{5}{6}}}{7^{\frac{5}{6}}} \\ \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=7^{\frac{5}{6}-\frac{5}{6}} \\ \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=7^0 \end{gathered}[/tex]
By definition
[tex]\begin{gathered} a^0=1 \\ \text{Where a }\ne0 \end{gathered}[/tex]
So
[tex]\begin{gathered} \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=7^0 \\ \frac{\sqrt[3]{7}\sqrt[]{7}}{\sqrt[6]{7^5}}=1 \end{gathered}[/tex]
Therefore, the correct answer is b. 1.
