Answer
Verified
439.8k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE