Question

राजा हरिश्चंद्र च पुर्ण नाव​

Answer 1

Answer:

I hope is helpful

Explanation:

Raja Harishchandra Bos

Mark I am marathi

Answer 2

Given :-

$\qquad$ ☀️$\sf x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\qquad$ ☀️$\sf y = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

To Find :-

$\qquad$ ☀️$\sf {x}^{2} + {y}^{2} = ?$

Solution :-

$\sf\underline{{Rationalising :- }} \\\\\green{ \qquad\leadsto\quad\sf x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }}$

$\qquad\leadsto\quad\sf x = \frac{ (\sqrt{3} + \sqrt{2} )^{2} }{ {( \sqrt{3}})^{2} - { (\sqrt{2} })^{2} }$

$\qquad\leadsto\quad\sf x = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} + 2 \sqrt{6} }{3 - 2}$

$\qquad\leadsto\quad\sf x =3 + 2 + 2 \sqrt{6}$

$\green{\qquad\leadsto\quad\sf x =5 + 2 \sqrt{6} }$

$\qquad\leadsto\quad \sf\underline{{Similarly : -}} \\\\ \purple{\qquad\leadsto\quad\sf y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }}$

$\qquad\leadsto\quad\sf y = \frac{ (\sqrt{3} - \sqrt{2})^{2} }{ {( \sqrt{3} )}^{2} - { (\sqrt{2} })^{2} }$

$\qquad\leadsto\quad\sf y= \frac{ { (\sqrt{3} })^{2} + {( \sqrt{2} )}^{2} - 2 \sqrt{6} }{3 - 2 }$

$\qquad\leadsto\quad\sf y=3 + 2 - 2 \sqrt{6}$

$\purple{\qquad\leadsto\quad\sf y =5 - 2 \sqrt{6}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad\leadsto\quad\sf\underline{{ Putting \: values : -}} \\\\\pink{\qquad\leadsto\quad\sf {x}^{2} + {y}^{2} =(x + y)^{2} - 2xy }\\ \\ \sf \leadsto {x}^{2} + {y}^{2} =(5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} )^{2} - 2(5 + 2\sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ \sf \leadsto {x}^{2} + {y}^{2} = {10}^{2} - 2( {5}^{2} - {(2 \sqrt{6}) }^{2} \\ \\ \sf \leadsto {x}^{2} + {y}^{2} =100 - 2(25 - 24) \\ \\ \sf \leadsto {x}^{2} + {y}^{2} =100 - 2 \\ \\ \pink{\sf \leadsto {x}^{2} + {y}^{2} =98}$