Question

plz solve anyone.I can't solve plz anyone​

Answer 1

2b   $\sqrt{a^2 + 2b\sqrt{a^2 - b^2} } + \sqrt{a^2 - 2b\sqrt{a^2 - b^2} } = 2b$

Step-by-step explanation:

Let say

$N = \sqrt{a^2 + 2b\sqrt{a^2 - b^2} } + \sqrt{a^2 - 2b\sqrt{a^2 - b^2} }$

Squaring both sides

$N^2 = (\sqrt{a^2 + 2b\sqrt{a^2 - b^2} })^2 + (\sqrt{a^2 - 2b\sqrt{a^2 - b^2} })^2 + 2 (\sqrt{a^2 + 2b\sqrt{a^2 - b^2} }) (\sqrt{a^2 - 2b\sqrt{a^2 - b^2} })$

$N^2 =a^2 + 2b\sqrt{a^2 - b^2 } + {a^2 - 2b\sqrt{a^2 - b^2} + 2 \sqrt{(a^2)^2 - (2b)^2(a^2 - b^2)}$

$N^2 =2a^2 + 2 \sqrt{(a^2)^2 - (2b)^2a^2 + (2b)^2(b^2)}\\N^2 =2a^2 + 2 \sqrt{(a^2)^2 - 2(2b^2)a^2 + (2b^2)^2} \\N^2 =2a^2 + 2 \sqrt{(2b^2 - a^2)^2} \\N^2 =2a^2 + 2(2b^2 - a^2) \\ N^2 =2a^2 + 4b^2 - 2a^2 \\N^2 = 4b^2$

Taking square root both side

N = 2b

Hence

$\sqrt{a^2 + 2b\sqrt{a^2 - b^2} } + \sqrt{a^2 - 2b\sqrt{a^2 - b^2} } = 2b$

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Answer 2

Solution

$\sqrt{ {a}^{2} + 2b \sqrt{ {a}^{2} - {b}^{2}} } + \sqrt{ {a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2} } }$

$\text{let } x = \sqrt{ {a}^{2} + 2b \sqrt{ {a}^{2} - {b}^{2}} } + \sqrt{ {a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2}} }$

Squaring on both sides

$\implies {x}^{2} = \bigg( \sqrt{ {a}^{2} + 2b \sqrt{ {a}^{2} - {b}^{2} } } + \sqrt{ {a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2}} } \bigg)^{2}$

$\implies {x}^{2} = \bigg(\sqrt{{a}^{2} + 2b \sqrt{{a}^{2} - {b}^{2}}} \bigg)^{2}+\bigg(\sqrt{ {a}^{2} - 2b \sqrt{{a}^{2} - {b}^{2}}} \bigg)^{2}\\ \\ + 2 \bigg(\sqrt{{a}^{2} + 2b\sqrt{{a}^{2} - {b}^{2}}}\bigg) \bigg(\sqrt{ {a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2} } } \bigg)$

[ Because (p + q)² = p² + q² + 2pq ]

$\implies {x}^{2} = {a}^{2} + 2b \sqrt{ {a}^{2} - {b}^{2} } + {a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2} }+ 2 \bigg( \sqrt{ ({a}^{2} + 2b \sqrt{ {a}^{2} - {b}^{2}})({a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2}})} \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ ({a}^{2} + 2b \sqrt{ {a}^{2} - {b}^{2}})({a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2}})} \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(a^{2} )}^{2} - (2b \sqrt{ {a}^{2} - {b}^{2} } )^{2} } \bigg)$

[ Because (p + q)(p - q) = p² - q² ]

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(a^{2} )}^{2} - \{(2b) ^{2} (\sqrt{ {a}^{2} - {b}^{2} } )^{2} \} } \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(a^{2} )}^{2} - \{(4b^{2} ({a}^{2} - {b}^2 ) \} } \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(a^{2} )}^{2} - \{4 {a}^{2} b^{2} - 4{b}^4 ) \} } \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(a^{2} )}^{2} - 4 {a}^{2} b^{2} + 4{b}^4 ) \} } \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(a^{2} )}^{2} - 2({a}^{2} )(2b^{2}) + (2{b}^2)^{2} } \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \bigg( \sqrt{ {(2b^{2} )}^{2} - 2(2{b}^{2} )(a^{2}) + (a^2)^{2} } \bigg)$

$\implies {x}^{2} = 2a^2 + 2 \{\sqrt{ {(2b^{2} - {a}^{2} ) }^{2} } \}$

[ Because p² - 2pq + q² = (p - q)² ]

$\implies {x}^{2} = 2a^2 + 2(2b^{2} - {a}^{2} )$

$\implies {x}^{2} = 2a^2 + 4b^{2} - 2{a}^{2}$

$\implies {x}^{2} = 4b^{2}$

$\implies x = \sqrt{ 4b^{2}}$

$\implies \boxed{x = 2b}$

Hence the answer is (C) 2b.