
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Reading materials:<br/>Material 1: When two non-zero algebraic expressions containing square roots are multiplied together, if their product does not contain square roots, then these two expressions are rationalized factors of each other. For example: $\sqrt{3} \times \sqrt{3} = 3$, $(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2}) = 6-2 = 4$. We call $\sqrt{3}$ a rationalized factor of $\sqrt{3}$, and $\sqrt{6}-\sqrt{2}$ a rationalized factor of $\sqrt{6}+\sqrt{2}$.<br/>Material 2: If the denominator of an algebraic expression contains square roots, usually the numerator and denominator can be multiplied by the rationalized factor of the denominator to eliminate the square roots in the denominator. This transformation is called rationalizing the denominator.<br/>For example: $\frac{1}{{\sqrt{3}}} = \frac{{1 \times \sqrt{3}}}{{\sqrt{3} \times \sqrt{3}}} = \frac{{\sqrt{3}}}{3}$, $\frac{8}{{\sqrt{6}-\sqrt{2}}} = \frac{{8(\sqrt{6}+\sqrt{2})}}{{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}} = \frac{{8(\sqrt{6}+\sqrt{2})}}{4} = 2\sqrt{6}+2\sqrt{2}$.<br/>Please explore and solve the following problems following the methods in the materials.<br/>$(1)$ The rationalized factor of $\sqrt{13}$ is ______, and the rationalized factor of $\sqrt{7}+\sqrt{5}$ is ______. (Write one for each)<br/>$(2)$ Rationalize the denominators of the following expressions. (Requirements: Write out the transformation process)<br/>① $\frac{3}{{\sqrt{15}}}$; ② $\frac{{11}}{{2\sqrt{5}-3}}$.<br/>$(3)$ Calculate: $\frac{1}{{1+\sqrt{2}}}+\frac{1}{{\sqrt{2}+\sqrt{3}}}+\frac{1}{{\sqrt{3}+\sqrt{4}}}+…+\frac{1}{{\sqrt{2018}+\sqrt{2019}}}$.

Present the answer in LaTex format: \boxed{Your answer}
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