Question

Guys please help and please give the solution not just the answer❤️​

Answer 1

Answer:

$\frac{3 + \sqrt{5} }{3 - \sqrt{5} } + \frac{3 - \sqrt{5} }{3 + \sqrt{5} }$

$\frac{(3 + \sqrt{ 5}) {}^{2} + (3 - \sqrt{5} ) {}^{2} }{(3 - \sqrt{5}) (3 + \sqrt{5}) }$

$\frac{9 + 5 + 6 \sqrt{5} + 9 + 5 - 6 \sqrt{5} }{(3) {}^{2} - ( \sqrt{5}) {}^{2} }$

$\frac{28}{9 - 5} = \frac{28}{4} = \frac{14}{2} = 7$

Answer 2

$\tt \: \frac{ 3+ \sqrt{ 5 } }{ 3- \sqrt{ 5 } } + \frac{ 3- \sqrt{ 5 } }{ 3+ \sqrt{ 5 } }$

$\tt \: \frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+\frac{3-\sqrt{5}}{3+\sqrt{5}}$

• Consider$\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right).$ Multiplication can be transformed into difference of squares using the rule:$\tt \: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.$

$\tt \: \frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{3^{2}-\left(\sqrt{5}\right)^{2}}+\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ \\ \tt \: \frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{9-5}+\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ \\ \tt \: \frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}+\frac{3-\sqrt{5}}{3+\sqrt{5}}\\ \\ \tt \: Use \: binomial \: theorem \\ \\ \tt \: \left(a+b\right)^{2}=a^{2}+2ab+b^{2} \: to \: expand \: \left(3+\sqrt{5}\right)^{2}.\\ \\ \tt \: \frac{9+6\sqrt{5}+\left(\sqrt{5}\right)^{2}}{4}+\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ \\ \tt \: \frac{9+6\sqrt{5}+5}{4}+\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ \\ \tt \: \frac{14+6\sqrt{5}}{4}+\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ \\ \tt \: \frac{14+6\sqrt{5}}{4}+\frac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)} \\ \\ \tt \:$

• Consider $\tt \: \left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right).$Multiplication can be transformed into difference of squares using the rule: $\tt \: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.$

$\tt \: \frac{14+6\sqrt{5}}{4}+\frac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{3^{2}-\left(\sqrt{5}\right)^{2}} \\ \\ \tt \: \frac{14+6\sqrt{5}}{4}+\frac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{9-5} \\ \\ \tt \:\frac{14+6\sqrt{5}}{4}+\frac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4} \\ \\ \tt \: \frac{14+6\sqrt{5}}{4}+</p><p>\frac{\left(3-\sqrt{5}\right)^{2}}{4} \\ \\ \tt \: Use \: binomial \: theorem \\ \\ \tt \: \left(a-b\right)^{2}=a^{2}-2ab+b^{2} \: to \: expand \: \left(3-\sqrt{5}\right)^{2}.\\ \\ \tt \: \frac{14+6\sqrt{5}}{4}+\frac{9-6\sqrt{5}+5}{4} \\ \\ \tt \: \frac{14+6\sqrt{5}}{4}+\frac{14-6\sqrt{5}}{4} \\ \\ \tt \: \frac{14+6\sqrt{5}+14-6\sqrt{5}}{4} \\ \\ \tt \: \frac{28}{4} \\ \\ \tt \: 7$

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$\sf \: @WildCat7083$