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sladkih [1.3K]
1 year ago
12

Rationalize the denominator: GRADE 9 CBSE

Mathematics
1 answer:
kakasveta [241]1 year ago
7 0

Answer:

1. a)

\begin{gathered} \frac{2}{ \sqrt{3} - 1 } = \frac{2}{ \sqrt{3} - 1} \times \ \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } \\ = \frac{2 \sqrt{3} + 2 } {( \sqrt{3} ) {}^{2} - 1 {}^{2} } = \frac{2 \sqrt{} 3}{3 - 1} = \frac{2 \sqrt{3} }{2} \\ = \sqrt{3} \end{gathered}

\\

b)

\begin{gathered} = > \: \: \frac{7}{ \sqrt{12} - \sqrt{5} } \\ \\  = > \: \: \frac{7}{ \sqrt{12} - \sqrt{5} } \times \frac{ \sqrt{12} + \sqrt{5} }{ \sqrt{12} + \sqrt{5} } \\ \\  = > \: \: \frac{7( \sqrt{12} + \sqrt{5} ) }{ {( \sqrt{12}) }^{2} - {( \sqrt{5}) }^{2} } \\ \\ =  > \: \: \frac{7( \sqrt{12} + \sqrt{5}) }{12 - 5} \\ \\ => \: \: \frac{ \cancel{7}( \sqrt{12} + \sqrt{5} ) }{ \cancel{7}} \\ \\ => \: \: \sqrt{12} + \sqrt{5} \end{gathered}

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