Math, asked by lataji, 1 year ago

Pls solve this question

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Answers

Answered by sprao534
1

Please see the attachment

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lataji: May I know how did u got 1/root 2..?
sprao534: multiply and divide with Root 2
lataji: But are u sure because we multiply or divide something on both lhs and rhs but here no equal to is given...
sprao534: i multiplied numerator and denominator by v2. here there is no left hand side and right hand side.
sprao534: we can multiply and divide numerator and denominator with same nonzero number, there is no change in value
lataji: Ok got it thanks much
Answered by siddhartharao77
1

Answer:

√6

Step-by-step explanation:

Given:\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}

=>\sqrt{\frac{2(2+ \sqrt{3})}{2}}+\sqrt{\frac{2(2-\sqrt{3})}{2} }

=>\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}

=>\sqrt{\frac{3+1+2\sqrt{3}}{2}}+\sqrt{\frac{3+1-2\sqrt{3}}{2}}

=>\sqrt{\frac{(\sqrt{3})^2+(1)^2+2\sqrt{3}}{2}}+\sqrt{\frac{(\sqrt{3})^2+(1)^2 - 2\sqrt{3}}{2}}

=>\sqrt{\frac{(\sqrt{3}+1)^2}{2}}+\sqrt{\frac{(\sqrt{3}-1)^2}{2}}

=>\frac{\sqrt{3} + 1}{\sqrt{2}}+\frac{\sqrt{3} - 1}{\sqrt{2}}

=>\frac{2\sqrt{3}}{\sqrt{2}}

=>\sqrt{2} \sqrt{3}

=> \sqrt{6}

Hope it helps!

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