Question

plz solve my math question plz try in least time
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Answer 1

Question:

$\displaystyle \sf Simplify: \ \ \dfrac{1}{\sqrt{19 - \sqrt{360}}} \ - \ \dfrac{1}{\sqrt{21 - \sqrt{440}}} \ + \ \dfrac{2}{\sqrt{20 + \sqrt{396}}}$

Solution:

We'll take each part of the question and rationalize it, then calculate it.

$\sf Rationalizing: \ \ \dfrac{1}{\sqrt{19 - \sqrt{360}}}\\ \\ \\\sf \Longrightarrow \dfrac{1}{\sqrt{19 - \sqrt{360}}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{1}{\sqrt{19 - \sqrt{360}}} \times \dfrac{\sqrt{19 + \sqrt{360}}}{\sqrt{19 + \sqrt{360}}}\\ \\ \\ \\\sf Using \ the \ identity \ a^2 - b^2 = (a-b)(a+b)\\ \\ \\ \\\sf \Longrightarrow \dfrac{\sqrt{19 + \sqrt{360}}}{\sqrt{\left(19\right)^2 - \left(\sqrt{360}\right)^2}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{\sqrt{19 + \sqrt{360}}}{\sqrt{361 - 360}}$

$\Longrightarrow \sf \dfrac{\sqrt{19 + \sqrt{360}}}{\sqrt{1}}\\ \\ \\ \\\sf \Longrightarrow \sqrt{19 + \sqrt{360}}\\ \\ \\ \\\sf \Longrightarrow \sqrt{10 + 9 + \sqrt{360}}\\ \\ \\\sf Taking \ \sqrt{10} \ to \ be \ 'a' \ and \ \sqrt{9} \ to \ be \ 'b'.\\ \\ Using \ (a+b)^2 = a^2 + 2ab + b^2\ we \ get:\\ \\ \\ \\\sf \Longrightarrow \sqrt{\left(\sqrt{10} + \sqrt{9}\right)^2}\ \ \ \longmapsto \textcircled{\scriptsize \sf 1}$

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$\sf Rationalizing: \ \ \dfrac{1}{\sqrt{21 - \sqrt{440}}}\\ \\ \\\sf \Longrightarrow \dfrac{1}{\sqrt{21 - \sqrt{440}}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{1}{\sqrt{21 - \sqrt{440}}} \times \dfrac{\sqrt{21 + \sqrt{440}}}{\sqrt{21 + \sqrt{440}}}\\ \\ \\ \\\sf Using \ the \ identity \ a^2 - b^2 = (a-b)(a+b)\\ \\ \\ \\\sf \Longrightarrow \dfrac{\sqrt{21 + \sqrt{440}}}{\sqrt{\left(21\right)^2 - \left(\sqrt{440}\right)^2}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{\sqrt{21 + \sqrt{440}}}{\sqrt{441 - 440}}$

$\sf \Longrightarrow \dfrac{\sqrt{21 + \sqrt{440}}}{1}\\ \\ \\ \\\sf \Longrightarrow \sqrt{21 + \sqrt{440}}\\ \\ \\ \\\sf \Longrightarrow \sqrt{10 + 11 + \sqrt{440}}\\ \\ \\ \\Taking \ \sqrt{10} \ to \ be \ 'a' \ and \ \sqrt{11} \ to \ be \ 'b'.\\ \\Using \ (a+b)^2 = a^2 + 2ab + b^2 \ we \ get:\\ \\ \\ \\\sf \Longrightarrow \sqrt{\left(\sqrt{10} + \sqrt{11}\right)^2} \ \ \ \ \ \longmapsto \textcircled{\scriptsize \sf 2}$

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$\sf Rationalizing: \ \ \dfrac{2}{\sqrt{20 + \sqrt{396}}}\\ \\ \\\sf \Longrightarrow \dfrac{2}{\sqrt{29 + \sqrt{396}}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{2}{\sqrt{20 + \sqrt{396}}} \times \dfrac{\sqrt{20 - \sqrt{396}}}{\sqrt{20 - \sqrt{396}}}\\ \\ \\ \\\sf Using \ the \ identity \ a^2 - b^2 = (a-b)(a+b)\\ \\ \\ \\\sf \Longrightarrow \dfrac{2\left(\sqrt{20 - \sqrt{396}}\right)}{\sqrt{\left(20\right)^2 - \left(\sqrt{396}\right)^2}}$

$\sf \Longrightarrow \dfrac{2\left(\sqrt{20 - \sqrt{396}} \ \right)}{\sqrt{400-396}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{2 \left(\sqrt{20 - \sqrt{396}} \ \right)}{\sqrt{4}}\\ \\ \\ \\\sf \Longrightarrow \dfrac{2\left(\sqrt{20 - \sqrt{396}}\ \right)}{2}\\ \\ \\ \\\sf \Longrightarrow \sqrt{20 - \sqrt{396}}$

$\Longrightarrow \sf \sqrt{11 + 9 - \sqrt{396}}\\ \\ \\ \\\sf Let \ \sqrt{11} \ stand \ for \ 'a' \ and \ \sqrt{9} \ stand \ for \ 'b'.\\ \\\sf Using \ (a - b)^2 = a^2 - 2ab + b^2 \ we \ get:\\ \\ \\ \\\sf \Longrightarrow \sqrt{\left(\sqrt{11} - \sqrt{9}\right)^2}\ \ \ \ \ \longmapsto \textcircled{\scriptsize \sf 3}$

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$\sf To \ Find: \ \ \dfrac{1}{\sqrt{19 - \sqrt{360}}} \ - \ \dfrac{1}{\sqrt{21 - \sqrt{440}}} \ + \ \dfrac{2}{\sqrt{20 + \sqrt{396}}}$

Substitute the values of 1, 2 and 3 above.

$\Longrightarrow \sf \sqrt{\left(\sqrt{10} + \sqrt{9}\right)^2} - \sqrt{\left(\sqrt{10} + \sqrt{11}\right)^2} + \sqrt{\left(\sqrt{11} - \sqrt{9}\right)^2}\\ \\ \\\Longrightarrow \sf \left(\sqrt{10} + \sqrt{9}\right) -\left(\sqrt{10} + \sqrt{11}\right)+ \left(\sqrt{11} - \sqrt{9}\right)\\ \\ \\\Longrightarrow \sf \sqrt{10} + \sqrt{9} - \sqrt{10} - \sqrt{11}\right)+ \sqrt{11} - \sqrt{9}\\ \\ \\\Longrightarrow \sf \ 0$

$\sf \therefore \ \dfrac{1}{\sqrt{19 - \sqrt{360}}} \ - \ \dfrac{1}{\sqrt{21 - \sqrt{440}}} \ + \ \dfrac{2}{\sqrt{20 + \sqrt{396}}} = 0$

Answer: 0.