Question

rationalise the denominators of these questions

Answer 1

Hey friend,
Here is the answer you were looking for:
$idenities \: used \: \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ \\ 1) \\ \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } = a - b \sqrt{77} \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} + \sqrt{7} } \times \frac{ \sqrt{11} - \sqrt{7} }{ \sqrt{11} - \sqrt{7} } \\ \\ = \frac{ {( \sqrt{11} )}^{2} + { \sqrt{7}) }^{2} - 2 \times \sqrt{11} \times \sqrt{7} }{ {( \sqrt{11}) }^{2} - {( \sqrt{7} )}^{2} } \\ \\ = \frac{11 + 7 - 2 \sqrt{77} }{11 - 7} \\ \\ = \frac{18 - 2 \sqrt{77} }{4} \\ \\ = \frac{9 - \sqrt{77} }{2} \\ \\ a = \frac{9}{2} \\ \\ b = \frac{1}{2} \\ \\ 2) \\ \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5} \\ \\ = \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} } \\ \\ = \frac{ {(4)}^{2} + {(3 \sqrt{5}) }^{2} + 2 \times 4 \times 3 \sqrt{5} }{ {(4)}^{2} - {(3 \sqrt{5}) }^{2} } \\ \\ = \frac{16 + 45 + 24 \sqrt{5} }{16 - 45} \\ \\ = \frac{60 + 24 \sqrt{5} }{ - 29} \\ \\ = \frac{ - 60 - 24 \sqrt{5} }{29} \\ \\ a = \frac{ - 60}{29} \\ \\ b = \frac{ - 24}{29} \\ \\ 3) \\ \frac{ \sqrt{5} - 2 }{ \sqrt{5} + 2 } = \frac{ \sqrt{5} + 2 }{ \sqrt{5} - 2 } \\ \\ \frac{ \sqrt{5} - 2 }{ \sqrt{5} + 2} \times \frac{ \sqrt{5} - 2 }{ \sqrt{5} - 2} = \frac{ \sqrt{5} + 2 }{ \sqrt{5} - 2} \times \frac{ \sqrt{5} + 2 }{ \sqrt{5} + 2} \\ \\ \frac{ {( \sqrt{5}) }^{2} + {(2)}^{2} - 2 \times \sqrt{5} \times 2}{ {( \sqrt{5}) }^{2} - {(2)}^{2} } = \frac{ {( \sqrt{5} )}^{2} + {(2)}^{2} + 2 \times \sqrt{5} \times 2 }{ {( \sqrt{5}) }^{2} - {(2)}^{2} } \\ \\ \frac{5 + 4 - 4 \sqrt{5} }{5 - 4} = \frac{5 + 4 + 4 \sqrt{5} }{5 - 4} \\ \\ 9 - 4 \sqrt{5} = 9 + 4 \sqrt{5} \\ \\ - 4 \sqrt{5} (is \: not \: equal \: to) 4 \sqrt{5} \\ \\ 4) \\ \frac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } \times \frac{7 + 3 \sqrt{5} }{7 + 3 \sqrt{5} } \\ \\ = \frac{ {(7)}^{2} + {(3 \sqrt{5}) }^{2} + 2 \times 7 \times 3 \sqrt{5} }{ {(7)}^{2} - {(3 \sqrt{5}) }^{2} } \\ \\ = \frac{49 + 45 + 42 \sqrt{5} }{49 - 45} \\ \\ = \frac{94 + 42}{4} \\ \\ = \frac{47 + 21}{2} \\ \\ 5) \\ \frac{5 + 2 \sqrt{3} }{7 - 4 \sqrt{3} } = 47a + b \sqrt{3} \\ \\ = \frac{5 + 2 \sqrt{3} }{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ = \frac{5 \times 7 + 5 \times 4 \sqrt{3} + 2 \sqrt{3} \times 7 + 2 \sqrt{3} \times 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ = \frac{35 + 20 \sqrt{3} +14 \sqrt{3} + 24 }{49 - 48} \\ \\ = 59 + 34 \sqrt{3} \\ \\ a = 1.25 (approx)\\ \\ b = 34$



Hope this helps!!!!

@Mahak24

Thanks...
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