Answer

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**Hint:**We will first eliminate the square root from the numerator and denominator of the first term. We know that $x$ can be written as $\sqrt{\left( {{x}^{2}} \right)}$ and we can write $\sqrt{\left( {{x}^{2}} \right)}$ as ${{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}$ we will express the first term in this form. We can now simplify the first term easily. We will write the term in the form of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ we can eliminate the square root from the second term. We will get the answer very easily.

**Complete step by step answer:**

We have $\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$,

We can rewrite the above equation as,

$\Rightarrow {{\left[ {{\left( \dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} \right)}^{2}} \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}}$

We will know solve the first term of the expression, we get,

$\begin{align}

& \Rightarrow {{\left[ {{\left( \dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} \right)}^{2}} \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}} \\

& \Rightarrow {{\left[ \left( \dfrac{{{\left( \sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2} \right)}^{2}}}{{{\left( \sqrt{\sqrt{5}+1} \right)}^{2}}} \right) \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}} \\

& \\

\end{align}$

We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we will apply this identity in numerator and we will get,

$\Rightarrow {{\left[ \dfrac{\left( \sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)} \right)}{{{\left( \sqrt{\sqrt{5}+1} \right)}^{2}}} \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}}$

We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ , we will get,

\[\begin{align}

& \Rightarrow {{\left[ \dfrac{\left( \sqrt{5}+2+\sqrt{5}-2+2\sqrt{5-4} \right)}{\sqrt{5}+1} \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}} \\

& \Rightarrow {{\left[ \dfrac{2\sqrt{5}+2}{\sqrt{5}+1} \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}} \\

\end{align}\]

Taking 2 common from the numerator, we will get,

\[\begin{align}

& \Rightarrow {{\left[ \dfrac{2\left( \sqrt{5}+1 \right)}{\sqrt{5}+1} \right]}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}} \\

& \Rightarrow {{2}^{\dfrac{1}{2}}}-\sqrt{3-2\sqrt{2}} \\

\end{align}\]

We know that ${{x}^{\dfrac{1}{2}}}=\sqrt{x}$

\[\Rightarrow \sqrt{2}-\sqrt{3-2\sqrt{2}}\]

We can write $\sqrt{3}={{1}^{2}}+{{\left( \sqrt{2} \right)}^{2}}$

\[\Rightarrow \sqrt{2}-\sqrt{{{1}^{2}}+{{\left( \sqrt{2} \right)}^{2}}-2\sqrt{2}}\]

We can clearly see that the second term is of the form ${{a}^{2}}+{{b}^{2}}-2ab$ so we can write it as ${{\left( a-b \right)}^{2}}$, we will get,

$\begin{align}

& \Rightarrow \sqrt{2}+\sqrt{{{\left( 1-\sqrt{2} \right)}^{2}}} \\

& \Rightarrow \sqrt{2}+1-\sqrt{2} \\

& \Rightarrow 1 \\

\end{align}$

Answer-$\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}=1$

**Note:**

An alternate method of solving this question is using the conjugate method where we multiply the numerator and denominator by conjugate of the numerator. But this method is very tedious and time taking. Whenever we get the question with square root it is better to eliminate the square roots by writing the in the form $x=\sqrt{{{\left( x \right)}^{2}}}$ or try to express the expression in the form of identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ or ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ it will help to simplify the expression and we can get the answer easily.

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