Question

V 3+v2
V3-v2


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Answer 1

${~~~~\implies ~~~ \sf \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } }$

• On rationalizing the denominator term

${~~~~\implies ~~~ \sf \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } }$

${ ~~~~\implies ~~~\sf \dfrac{ {(\sqrt{3} + \sqrt{2})}^{2} }{{ ({\sqrt{3} )}^{2} - ( \sqrt{2} )}^{2} } }$

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

${ ~~~~\implies ~~~\sf \dfrac{ 3+ 2 + 2 \sqrt{6} }{ 1} }$

${~~~~\implies ~~~ \boxed{\bf 5 + 2 \sqrt{6} }}$

${~~~ ~~~ \sf \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \bf= 5 + 2 \sqrt{6} }$

• $\begin{gathered}\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\gray{\begin{gathered}\tiny\begin{gathered}\small{\small{\small{\small{\small{\small{\small{\small{\small{\small{\begin{gathered}\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\red{ \bigstar} \: \underline{\bf{\orange{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\\\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \\\dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab )\\\\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}}}}}}}}}}\end{gathered}\end{gathered}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\\ \end{gathered}$

Answer 2

*＊✿❀ Question ❀✿＊*

★ $\sqrt{3} - \: \sqrt{2} \\ \sqrt{3 \: } \: + \: \sqrt{2}$

*＊✿❀ Answer ❀✿＊*

➡ Solution in the attachment photo.

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✳ Mark as brainlist.