Question :- Simplify
[tex]\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ [/tex]
Answer:
[tex]\boxed{ \sf{ \:\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} = 0 \: }} \\ \\ [/tex]
Step-by-step explanation:
Given expression is
Consider,
[tex]\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: = \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \dfrac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{ (\sqrt{6})^{2} - (\sqrt{3}) ^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{6 - 3} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{3} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{2} ( \sqrt{6} + \sqrt{3} ) \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{12} + \sqrt{6} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } = \sqrt{12} + \sqrt{6} \\ \\ [/tex]
Now, Consider
[tex]\sf \: \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \times \dfrac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2}) }{( \sqrt{6})^{2} - (\sqrt{2}) ^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2}) }{6 - 2} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2}) }{4} \\ \\ [/tex]
[tex]\sf \: = \:\sqrt{3} ( \sqrt{6} + \sqrt{2}) \\ \\ [/tex]
[tex]\sf \: = \:\sqrt{18} + \sqrt{6} \\ \\ [/tex]
[tex]\sf\implies \sf \: \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } = \sqrt{18} + \sqrt{12} \\ \\ [/tex]
[tex]\sf \: \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \times \dfrac{ \sqrt{6} - 2 }{ \sqrt{6} - 2} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \sqrt{3} ( \sqrt{6} - 2)}{ (\sqrt{6})^{2} - {2}^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 ( \sqrt{18} - 2 \sqrt{3} )}{ 6 - 4} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 ( \sqrt{18} - \sqrt{12} )}{2} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{18} - \sqrt{12} \\ \\ [/tex]
[tex]\sf\implies \sf \: \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} = \sqrt{18} - \sqrt{12} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{12} + \sqrt{6} - \sqrt{18} - \sqrt{6} + \sqrt{18} - \sqrt{12} \\ \\ [/tex]
[tex]\sf \: = \:0 \\ \\ [/tex]
Hence
[tex]\sf\implies \sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} = 0 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
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Answers & Comments
Question :- Simplify
[tex]\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ [/tex]
Answer:
[tex]\boxed{ \sf{ \:\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} = 0 \: }} \\ \\ [/tex]
Step-by-step explanation:
Given expression is
[tex]\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ [/tex]
Consider,
[tex]\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: = \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } \times \dfrac{ \sqrt{6} + \sqrt{3} }{ \sqrt{6} + \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{ (\sqrt{6})^{2} - (\sqrt{3}) ^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{6 - 3} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{3 \sqrt{2} ( \sqrt{6} + \sqrt{3} )}{3} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{2} ( \sqrt{6} + \sqrt{3} ) \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{12} + \sqrt{6} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } = \sqrt{12} + \sqrt{6} \\ \\ [/tex]
Now, Consider
[tex]\sf \: \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: = \: \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \times \dfrac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2}) }{( \sqrt{6})^{2} - (\sqrt{2}) ^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2}) }{6 - 2} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2}) }{4} \\ \\ [/tex]
[tex]\sf \: = \:\sqrt{3} ( \sqrt{6} + \sqrt{2}) \\ \\ [/tex]
[tex]\sf \: = \:\sqrt{18} + \sqrt{6} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } = \sqrt{18} + \sqrt{12} \\ \\ [/tex]
Now, Consider
[tex]\sf \: \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: = \: \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \times \dfrac{ \sqrt{6} - 2 }{ \sqrt{6} - 2} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 \sqrt{3} ( \sqrt{6} - 2)}{ (\sqrt{6})^{2} - {2}^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 ( \sqrt{18} - 2 \sqrt{3} )}{ 6 - 4} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{2 ( \sqrt{18} - \sqrt{12} )}{2} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{18} - \sqrt{12} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} = \sqrt{18} - \sqrt{12} \\ \\ [/tex]
Now, Consider
[tex]\sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{12} + \sqrt{6} - \sqrt{18} - \sqrt{6} + \sqrt{18} - \sqrt{12} \\ \\ [/tex]
[tex]\sf \: = \:0 \\ \\ [/tex]
Hence
[tex]\sf\implies \sf \: \dfrac{3 \sqrt{2} }{ \sqrt{6} - \sqrt{3} } - \dfrac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } + \dfrac{2 \sqrt{3} }{ \sqrt{6} + 2} = 0 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]