First divide the roots of the same degree:

\(\sqrt[4]{48}:\sqrt[4]{3}=\sqrt[4]{\dfrac{48}{3}}=\sqrt[4]{16}=2\).

Now evaluate the power:

\(\left(\dfrac13\right)^{-2}=3^2=9\).

So the whole expression equals:

\(2+9=11\).

Simplify the first fraction:

\(25^5=(5^2)^5=5^{10}\), so

\(\dfrac{5^{11}}{25^5}=\dfrac{5^{11}}{5^{10}}=5\).

Next:

\(\sqrt[3]{\dfrac{27}{8}}=\dfrac{\sqrt[3]{27}}{\sqrt[3]{8}}=\dfrac{3}{2}\).

Therefore,

\(5-\dfrac{3}{2}=\dfrac{10}{2}-\dfrac{3}{2}=\dfrac{7}{2}\).

Combine the roots of the same degree:

\(\sqrt[5]{8}\cdot\sqrt[5]{4}=\sqrt[5]{32}=2\).

Also, \(10^{-1}=\dfrac{1}{10}\).

Then

\(2-\dfrac{1}{10}=\dfrac{20}{10}-\dfrac{1}{10}=\dfrac{19}{10}\).

\(\sqrt[4]{81}=3\), since \(81=3^4\).

Next:

\(8^4=(2^3)^4=2^{12}\), so

\(\dfrac{2^{11}}{8^4}=\dfrac{2^{11}}{2^{12}}=2^{-1}=\dfrac{1}{2}\).

Therefore,

\(3-\dfrac{1}{2}=\dfrac{6}{2}-\dfrac{1}{2}=\dfrac{5}{2}\).

Use the difference of squares formula:

\((\sqrt{11}+\sqrt{3})(\sqrt{11}-\sqrt{3})=11-3=8\).

Then

\(\sqrt[3]{8}=2\).

First compute the first part:

\((3^{-2})^{-5}=3^{10}\).

Then

\(\dfrac{(3^{-2})^{-5}}{3^9}=\dfrac{3^{10}}{3^9}=3\).

In the second part, combine the cube roots:

\(\sqrt[3]{\dfrac12}\cdot\sqrt[3]{\dfrac14}=\sqrt[3]{\dfrac18}=\dfrac12\).

So,

\(3+\dfrac12=\dfrac{6}{2}+\dfrac12=\dfrac{7}{2}\).

Simplify the first fraction:

\(6^{10}=(2\cdot3)^{10}=2^{10}3^{10}\).

Then

\(\dfrac{6^{10}}{3^{10}\cdot2^8}=\dfrac{2^{10}3^{10}}{3^{10}2^8}=2^{2}=4\).

Next,

\(\left(\dfrac{10}{3}\right)^{-1}=\dfrac{3}{10}\).

Therefore,

\(4-\dfrac{3}{10}=\dfrac{40}{10}-\dfrac{3}{10}=\dfrac{37}{10}\).

\(27^{1/3}=3\).

\(\left(\dfrac12\right)^{-2}=2^2=4\).

Therefore,

\(3+4=7\).

Divide the cube roots:

\(\sqrt[3]{270}:\sqrt[3]{10}=\sqrt[3]{\dfrac{270}{10}}=\sqrt[3]{27}=3\).

Next,

\(\left(\dfrac15\right)^{-1}=5\).

So,

\(3+5=8\).

Since \(49^6=(7^2)^6=7^{12}\), we get

\(\dfrac{7^{13}}{49^6}=\dfrac{7^{13}}{7^{12}}=7\).

Next:

\(\sqrt[4]{\dfrac{625}{16}}=\dfrac{\sqrt[4]{625}}{\sqrt[4]{16}}=\dfrac{5}{2}\).

Therefore,

\(7-\dfrac{5}{2}=\dfrac{14}{2}-\dfrac{5}{2}=\dfrac{9}{2}\).

Combine the cube roots:

\(\sqrt[3]{25}\cdot\sqrt[3]{5}=\sqrt[3]{125}=5\).

Also, \(4^{-1}=\dfrac14\).

Therefore,

\(5-\dfrac14=\dfrac{20}{4}-\dfrac14=\dfrac{19}{4}\).

Since \(125=5^3\), we have

\(125^5=(5^3)^5=5^{15}\).

Therefore,

\(\dfrac{125^5}{5^{14}}=5\).

Also, \(\sqrt[6]{64}=2\), because \(64=2^6\).

So,

\(5+2=7\).

Apply the difference of squares formula:

\((\sqrt{30}+\sqrt{3})(\sqrt{30}-\sqrt{3})=30-3=27\).

Then

\(\sqrt[3]{27}=3\).

First compute the first part:

\((4^{-2})^{-6}=4^{12}\).

Then

\(\dfrac{(4^{-2})^{-6}}{4^{11}}=4\).

In the second part, combine the cube roots:

\(\sqrt[3]{\dfrac{1}{10}}\cdot\sqrt[3]{\dfrac{1}{100}}=\sqrt[3]{\dfrac{1}{1000}}=\dfrac{1}{10}\).

Therefore,

\(4+\dfrac{1}{10}=\dfrac{40}{10}+\dfrac{1}{10}=\dfrac{41}{10}\).

Simplify the first fraction:

\(10^{12}=(2\cdot5)^{12}=2^{12}5^{12}\).

Then

\(\dfrac{10^{12}}{5^{10}\cdot2^{12}}=5^{2}=25\).

Next,

\(\left(\dfrac{10}{7}\right)^{-1}=\dfrac{7}{10}\).

Therefore,

\(25-\dfrac{7}{10}=\dfrac{250}{10}-\dfrac{7}{10}=\dfrac{243}{10}\).

\(36^{1/2}=\sqrt{36}=6\).

Also, \(\left(\dfrac23\right)^{-1}=\dfrac32\).

Then

\(6+\dfrac32=\dfrac{12}{2}+\dfrac32=\dfrac{15}{2}\).