Question

pls help me with the problem of rational numbers

sole for a and b.


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Answer 1

$\frac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } = a \sqrt{5} - b \\ \\ = \frac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } \times \frac{3 - 2 \sqrt{5} }{3 - 2 \sqrt{5} } \\ \\ = \frac{(3 - \sqrt{5})(3 - 2 \sqrt{5} ) }{(3 + 2 \sqrt{5})(3 - 2 \sqrt{5)} } \\ \\ = \frac{3(3 - 2 \sqrt{5 } ) - \sqrt{5} (3 - 2 \sqrt{5} )}{ {3}^{2} - ({2 \sqrt{5} )}^{2} } \\ \\ = \frac{ {3}^{2} - 6 \sqrt{5} - 3 \sqrt{5} + 2(5)}{9 - 4(5)} \\ \\ = \frac{9 - 9 \sqrt{5} + 10}{9 - 20} \\ \\ = \frac{19 - 9 \sqrt{5} }{ - 11} \\ \\ = \frac{ - (19 - 9 \sqrt{5}) }{11} \\ \\ = \frac{ - 19 + 9 \sqrt{5} }{11} \\ \\ = \frac{9 \sqrt{5} - 19 }{11} \\ \\ = \frac{ 9 \sqrt{5} }{11} - \frac{19}{11} \\ \\ = a \sqrt{5} - \frac{19}{11} \\ \\ ∴, \: a = \frac{9}{11}$

if it is $\frac{a \sqrt{5} - 19}{11}$

then a = 9

But ATQ,

$\frac{a \sqrt{5} - 19}{11}$

So, $\frac{19}{11}$

#NAWABZAADI

Answer 2

Answer:

you can prefer above answer, suneo