Question

if k= √44+24√2 - √11-6√2 ÷ √2+1 , then what is the value of k/3​

Answer 1

Given : k=[√(44+24√2)+√(11-6√2)]/(√2+1)​    ( k = $\frac{\sqrt{44+24\sqrt{2} } - \sqrt{11-6\sqrt{2} } }{\sqrt{2} + 1}$ )

To find : k/3

Solution:

k=[√(44+24√2)+√(11-6√2)]/(√2+1)​  

k = $\frac{\sqrt{44+24\sqrt{2} } - \sqrt{11-6\sqrt{2} } }{\sqrt{2} + 1}$

44 + 24√2  

= 8 + 36 + 24√2

= (2√2)² + 6² + 2(2√2)6

= (6 + 2√2)²

=> [√(44+24√2) =  6 + 2√2

11-6√2

= 9 + 2 - 6√2

= 3² + √2²  - 2(3)(√2)

= ( 3 - √2)²

=> √(11-6√2) = (3 - √2)

√(44+24√2)  + √(11-6√2)  =   6 + 2√2  -  (3 - √2)

=> √(44+24√2)  + √(11-6√2)  = 3 + 3√2

=>  √(44+24√2)  + √(11-6√2)  = 3(1 + √2 )

k=[√(44+24√2)+√(11-6√2)]/(√2+1)​   = 3(1 + √2 ) / (√2+1)​

=> k = 3

=> k/3 = 1

k/3 = 1

Learn More:

k=[√(44+24√2)+√(11-6√2)]/(√2+1)​

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