Question

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Answer 1

Step-by-step explanation:

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Answer 2

Given

$\sf If\:x=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\:and\:y=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}$

To find

Find the value of x² + y² + xy

Solution

$\implies\sf If\;x=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$

★ Rationalize its denominator

$\implies\sf x=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\times\dfrac{\sqrt{2}+1}{\sqrt{2}+1} \\ \\ \implies\sf x=\dfrac{(\sqrt{2}+1)^2}{(\sqrt{2})^2-(1)^2} \\ \\ \implies\sf x=\dfrac{(\sqrt{2})^2+(1)^2+2\times\sqrt{2}\times{1}}{2-1} \\ \\ \implies\sf 2+1+2\sqrt{2}\\ \\ \implies\sf x=3+2\sqrt{2}$

__________________________

$\implies\sf If\;y=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}$

★ Rationalize its denominator

$\implies\sf y=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\times\dfrac{\sqrt{2}-1}{\sqrt{2}-1} \\ \\ \implies\sf y=\dfrac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-(1)^2} \\ \\ \implies\sf y=\dfrac{(\sqrt{2})^2+(1)^2-2\times\sqrt{2}\times{1}}{2-1} \\ \\ \implies\sf 2+1-2\sqrt{2} \\ \\ \implies\sf y=3-2\sqrt{2}$

Now, the value of x² + y² + xy

Putting the value of both x and y

$\implies\sf x^2+y^2+xy$

$\sf =(3+2\sqrt{2})^2+(3-2\sqrt{2})^2+(3+2\sqrt{2})(3-2\sqrt{2}) \\ \\ \sf =(3)^2+(2\sqrt{2})^2+2\times{3}\times{2\sqrt{2}}+$

$\sf (3)^2+(2\sqrt{2})^2-2\times{3}\times{2\sqrt{2}}+(3)^2-(2\sqrt{2})^2 \\ \\ \sf =9+8+12{\sqrt{2}}+9+8-12{\sqrt{2}}+9-8 \\ \\ \sf =17+17+1 \\ \\ \sf =35$