Question

✓7(✓5 - ✓2) - ✓5(✓7 - ✓2) + 2✓2 /✓5 +✓7​

Answer 1

Answer:

I think this is the answer please do confirm the answer from external sources.

Answer 2

$\sf= > \sqrt{7} ( \sqrt{5} - \sqrt{2} ) - \sqrt{5} ( \sqrt{7} - \sqrt{2} ) + \frac{2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \sqrt{7} \sqrt{5} - \sqrt{2} \sqrt{7} - \sqrt{5} \sqrt{7} + \sqrt{5} \sqrt{2} + \frac{2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{( \sqrt{5} + \sqrt{7})( \sqrt{7} \sqrt{5} - \sqrt{2} \sqrt{7}) - ( \sqrt{5} + \sqrt{7})( \sqrt{5} \sqrt{7} + \sqrt{5} \sqrt{2} ) + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{( \sqrt{5} + \sqrt{7} )( \sqrt{35} - \sqrt{14} ) - ( \sqrt{5} + \sqrt{7} )( \sqrt{35} + \sqrt{10}) + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{ \sqrt{5} ( \sqrt{35} - \sqrt{14} ) + \sqrt{7} ( \sqrt{35} - \sqrt{14} ) - \sqrt{5} ( \sqrt{35} + \sqrt{10}) + \sqrt{7} ( \sqrt{35} + \sqrt{10} ) + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{ \cancel{{ \sqrt{175}}} - { \cancel{\sqrt{70}}} + \sqrt{245} - \sqrt{98} - { \cancel{\sqrt{175}}} - \sqrt{50} + \sqrt{245} + { \cancel{\sqrt{70}}} + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{ \sqrt{245} - \sqrt{98} - \sqrt{50} + \sqrt{245} + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{7 \sqrt{5} - 7 \sqrt{2} - 5 \sqrt{2} + 7 \sqrt{5} + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{14 \sqrt{5} - 7 \sqrt{2} - 5 \sqrt{2} + 2 \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{1 4\sqrt{5} - (7 + 5 - 2) \sqrt{2} }{ \sqrt{5} + \sqrt{7} }$

$\sf= > \frac{14 \sqrt{5} - 10 \sqrt{2} }{ \sqrt{5} + \sqrt{7} } \times \frac{ \sqrt{5} - \sqrt{7} }{ \sqrt{5} - \sqrt{7} }$

$\sf= > \frac{14 \sqrt{5} ( \sqrt{5} - \sqrt{7} ) - 10 \sqrt{2} ( \sqrt{5} - \sqrt{7} )}{ {( \sqrt{5} )}^{2} - {( \sqrt{7} )}^{2} }$

Identity used here :

(a+b)(a-b)= a²-b²

$\sf= > \frac{14 \times 5 - 14 \sqrt{35} - 10 \sqrt{2} \sqrt{5} + 10 \sqrt{14} }{5 - 7}$

$\sf= > \frac{70 - 82.82 - 31.62 + 37.41}{ - 2}$

$\sf= > \frac{ - 12.82 + 5.79}{ - 2} = \frac{ - 7.03}{ - 2} = \bold{3.515}$