Question

under root 5 minus one upon under root 5 + 1 + under root 5 + 1 upon under root 5 - 1 rationalism it

Answer 1

Hey friend,
Here is answer you were looking for:
$\frac{ (\sqrt{5)} - 1 }{ (\sqrt{5}) + 1 } + \frac{( \sqrt{5}) + 1 }{( \sqrt{5}) - 1 } \\ \\ = \frac{( \sqrt{5} ) - 1}{( \sqrt{5}) + 1 } \times \frac{( \sqrt{5}) - 1 }{( \sqrt{5} ) - 1} + \\ \frac{( \sqrt{5} ) + 1}{( \sqrt{5} ) - 1} \times \frac{( \sqrt{5}) + 1 }{( \sqrt{5} ) + 1} \\ \\ = \frac{( \sqrt{5})^{2} + (1)^{2} + 2 \times ( \sqrt{5 )} \times 1 }{( \sqrt{5} )^{2} - (1)^{2} } \\ + \frac{( \sqrt{5})^{2} + {(1)}^{2} + 2 \times ( \sqrt{5}) \times 1 }{( \sqrt{5})^{2} - (1)^{2} } \\ \\ = \frac{5 + 1 - 2 \sqrt{5} }{5 - 1} + \frac{5 + 1 + 2 \sqrt{5} }{5 - 1} \\ \\ = \frac{6 - 2 \sqrt{5} + 6 + 2 \sqrt{5} }{4} \\ \\ = \frac{6 + 6}{4} \\ \\ = \frac{12}{4} \\ \\ = 3$

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@Mahak24

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Answer 2

Step-by-step explanation:

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