Question

plz answer fast...(+_+)​

Answer 1

Answer:

The correct Answer is:-

option ( A)

Answer 1

Answer 2

$\sf{If \: N = \frac{ \sqrt{ \sqrt{5}+2 } + \sqrt{ \sqrt{5}-2 } }{ \sqrt{ \sqrt{5} + 1 } } - \sqrt{3 - 2 \sqrt{2} }}$

$\sf{then\: N\: = }$

$\sf(a) = 1 \: \: \: \: \: \: \: \: \: (b) = 2 \sqrt{2} - 1 \\ \sf(c) = \frac{ \sqrt{5} }{2} \: \: \: \: \: \: \: \: \: (d) = \frac{2}{ \sqrt{ \sqrt{5} + 1 } }$

$\sf{Answer\: = \bold{1}}$

Step by step explanation:-

$\sf{\Rightarrow \frac{ \sqrt{ \sqrt{5}+2 } + \sqrt{ \sqrt{5}-2 } }{ \sqrt{ \sqrt{5} + 1 } } -\sqrt{3 - 2 \sqrt{2} }}$

$\sf{let \Rightarrow \frac{ \sqrt{ \sqrt{5}+2 } + \sqrt{ \sqrt{5}-2 } }{ \sqrt{ \sqrt{5} + 1 } } \: \: \: \: be \: \: (i)}$

$\sf{and \Rightarrow \sqrt{3 - 2 \sqrt{2} } \: \: \: \: \: \: be \: \: \: \: \: (ii)}$

solving equation (i)

$\sf{\Rightarrow \frac{ \sqrt{ \sqrt{5}+2 } + \sqrt{ \sqrt{5}-2 } }{ \sqrt{ \sqrt{5} + 1 } }}$

Rational the denominator term

$\sf{\Rightarrow \frac{ \sqrt{ \sqrt{5}+2 } + \sqrt{ \sqrt{5}-2 } }{ \sqrt{ \sqrt{5} + 1 } } \times \frac{ \sqrt{ \sqrt{5} - 1} }{ \sqrt{ \sqrt{5} - 1} } }$

$\sf{\Rightarrow \frac{ (\sqrt{ \sqrt{5} + 2 })( \sqrt{ \sqrt{5} - 1 }) + ( \sqrt{ \sqrt{5} - 2 })( \sqrt{ \sqrt{5} - 1)} }{ \sqrt{ ( \sqrt{5})^{2} - {(1)}^{2} } }}$

$\sf{\Rightarrow \frac{ \sqrt{5 - \sqrt{5} + 2 \sqrt{5} - 2 } + \sqrt{5 - \sqrt{5} - 2 \sqrt{5} + 2 } }{\sqrt{5 - 1} }}$

$\sf{\Rightarrow \frac{ \sqrt{5 + \sqrt{5} - 2 } + \sqrt{5 - 3\sqrt{5} + 2 } }{\sqrt{4} }}$

$\sf{\Rightarrow \frac{ \sqrt{3 + \sqrt{5} } + \sqrt{7 - 3\sqrt{5} } }{\cancel{\sqrt{ {2}^{\cancel2} }} }}$

$\sf{\Rightarrow \frac{ \sqrt{(3 + \sqrt{5} )^{ \frac{2}{2} } } + \sqrt{(7 - 3\sqrt{5})^{ \frac{2}{2} } } }{2}}$

$\sf{\Rightarrow \frac{ \sqrt{ 6 + 2\sqrt{5} } + \sqrt{14- 6 \sqrt{5} } }{{2\sqrt{ {2} }} }}$

$\sf{\Rightarrow \frac{ \sqrt{ 1 + 5 + 2\sqrt{5} } + \sqrt{9 + 5- 6 \sqrt{5} } }{{2\sqrt{ {2} }} }}$

$\sf{\Rightarrow \frac{ \sqrt{ (1 + \sqrt{5})^{2} } + \sqrt{(3 - \sqrt{5})^{2} } }{{2\sqrt{ {2} }} }}$

$\sf{\Rightarrow \frac{\cancel{ \sqrt{ (1 + \sqrt{5})^{\cancel2} } } + \cancel{\sqrt{(3 - \sqrt{5})^{\cancel2} }} }{{2\sqrt{ {2} }} }}$

$\sf{\Rightarrow \: \frac{1 + \sqrt{5} + 3 - \sqrt{5} }{2 \sqrt{2} } }$

$\sf{\Rightarrow \: \frac{1 + \cancel{\sqrt{5}} + 3 - \cancel{ \sqrt{5}} }{2 \sqrt{2} } }$

$\sf{\Rightarrow \: \frac{4}{2 \sqrt{2} } }$

$\sf{\Rightarrow \: \frac{\cancel{4}^{2} } {\cancel2 \sqrt{2} } }$

$\sf{\Rightarrow \frac{ \sqrt{2} \times \sqrt{2} }{ \sqrt{2} }}$

$\sf{\Rightarrow \: \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: : \:(i)}$

solving equation (ii)

$\sf{\Rightarrow \sqrt{3 - 2\sqrt{2} } }$

$\sf{\Rightarrow \sqrt{1 + 2 - 2\sqrt{2} } }$

$\sf{\Rightarrow \sqrt{ (\sqrt{2} - 1 ) ^{2} \: \: \: }}$

$\sf{\Rightarrow \: \sqrt{2} - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (ii)}$

On subtracting (ii) from (i) we get

$\sf{\Rightarrow\sqrt{2} - ( \sqrt{2} - 1)}$

$\sf{\Rightarrow \sqrt{2} - \sqrt{2} + 1}$

$\sf{\Rightarrow\cancel{ \sqrt{2} } - \cancel{ \sqrt{2} } + 1}$

$\sf{\Rightarrow1}$

For More Information

• $(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab$
• $(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab$
• $(a + b)(a - b) = {a}^{2} - {b}^{2}$
• $(a + b + c)^{2} = {a}^{2} + {b}^{2} + {c}^{2} +2ab + 2bc + 2ca$
• $(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b)$
• $(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b)$
• $a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab)$
• $a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab)$