Question

what if the value of √162+√108 /√72+√48?

Answer 1

HELLO....FRIEND!!

THE ANSWER IS HERE,

$= > \: \frac{ \sqrt{162} + \sqrt{108} }{ \sqrt{72} + \sqrt{48} }$

$= > \: \frac{9 \sqrt{2} + 6 \sqrt{3} }{4 \sqrt{3} + 6 \sqrt{2} }$


$= > \: \frac{9 \sqrt{2} + 6 \sqrt{3} }{4 \sqrt{3} + 6 \sqrt{2} } \times \frac{6 \sqrt{2} - 4 \sqrt{3} }{6 \sqrt{2} - 4 \sqrt{3} }$

$= > \: \frac{108 - 36 \sqrt{6} + 36 \sqrt{6} - 72}{72 - 48}$

$= > \: \frac{36}{24}$

$= > \: \frac{3}{2}$


:-)Hope it helps u.

Answer 2

$\frac{ \sqrt{162} + \sqrt{108} }{\sqrt{72} + \sqrt{48} }$
$\frac{ \sqrt{2 \times 3 \times 3 \times 3 \times 3} + \sqrt{2 \times 2 \times 3 \times 3 \times 3} }{ \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times } + \sqrt{2 \times 2 \times 2 \times 2 \times 3 \} } }$
$\frac{3 \times 3 \sqrt{2} + 2 \times 3 \sqrt{3} }{2 \times 3 \sqrt{2} + 2 \times 2 \sqrt{3} }$
$\frac{9 + 6}{6+ 4}$
$\frac{15}{10}$
$\frac{3}{2}$