Question

THE VALUE OF THE ABOVE QUESTION IS EQUALS TO?​

Answer 1

Answer:

PROCEDURE

We can solve this question by firstly breaking the numerator in its simplest form.

After breaking we will take three common from each other and then left will be cancelled by the denominator and the final answer came 3 as:-)

We can write

$\sqrt{27} \: as \: 3 \sqrt{3} \\ \sqrt{45}\: as \: 3 \sqrt{5} \\ \sqrt{63} \: as \: 3 \sqrt{7}$

Now:

$\frac{ \sqrt{27} + \sqrt{45} + \sqrt{63} }{ \sqrt{3 } + \sqrt{5} + \sqrt{7} } \\ = \frac{3 \sqrt{3} + 3 \sqrt{5} + 3 \sqrt{7} }{ \sqrt{3} + \sqrt{5} + \sqrt{7} } \\ = \frac{3( \cancel{ \sqrt{3} + \sqrt{5} + \sqrt{7} }}{ \cancel{ \sqrt{ 3} + \sqrt{5} + \sqrt{7} }} \\ = 3$

Answer 2

Answer:

3

Step-by-step explanation:

$\sqrt{27} = 3\sqrt{3} ; \sqrt{45} = 3\sqrt{5} ; \sqrt{63} = 3\sqrt{7}$

$\sqrt{27}+\sqrt{45}+\sqrt{63} = 3(\sqrt{3} + \sqrt{5} + \sqrt{7})$

$3(\sqrt{3} + \sqrt{5} + \sqrt{7}) / \sqrt{3} + \sqrt{5} + \sqrt{7} = 3$

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