Question

help I will mark the brainliset pls ​

Answer 1

Answer:

So=35.Hence RHS=LHS.

I have proved.

check and tell mee

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak {\: \: Given, \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \mathtt{x = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1} } \\ \\ \: \: \: \: \: \: \: \: \: \mathtt{y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1 } } \\ \\ \\ \underline{ \mathfrak{ \: \: To \: \: find : \mapsto \: }} \\ \\ \: \: \: \: \: \: \: \: \mathtt{The \: \: value \: \: of \: \: \: \: \red{x {}^{2} + y {}^{2} } }$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: }} \\ \\ \: \: \: \: \: \: \: \: \mathtt{ (x + y) {}^{2} = x {}^{2} +2xy + y {}^{2} } \\ \\ \mathtt{ : \rightsquigarrow : \:(x + y) {}^{2} - 2xy = x {}^{2} + \cancel{2xy} + y {}^{2} - \cancel{2xy}} \\ \\ \mathtt{ : \rightsquigarrow : \:(x + y) {}^{2} - 2xy = x {}^{2} + y {}^{2} } \\ \\ \mathtt{ : \rightsquigarrow : \: \underline{ \: \: x {}^{2} + y {}^{2} = (x + y) {}^{2} - 2xy \: \: }}$

$\underline{ \mathfrak{ \: \: Now, \: \: }} \\ \\ \: \: \: \: \: \: \mathtt{x + y = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 } + \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = \frac{( \sqrt{2} + 1) {}^{2} + ( \sqrt{2} - 1) {}^{2} }{( \sqrt{2} - 1)( \sqrt{2} + 1) }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{= \frac{(2 + 2 \sqrt{2} + 1) + (2 - 2 \sqrt{2} + 1) }{( \sqrt{2} ){}^{2} - (1) {}^{2} } } \\ \\\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt {= \frac{2 + 2 \sqrt{2} + 1 + 2 - 2 \sqrt{2} + 1 }{2 - 1} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = \frac{6}{1} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{= 6} \\ \\ \: \: \: \: \: \mathtt{ \therefore \: \underline{ \: \: ( x + y) {}^{2} = (6) {}^{2} = 36 \: \: } }$

$\underline{ \mathfrak{ \: \: Again, \: \: }} \\ \\ \mathtt{xy = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 } \times \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} } \\ \\ \: \: \: \: \: \: \mathtt {= 1} \\ \\ \mathtt{ \therefore \: \: \underline{ \: \: 2xy = 2 \times 1 = 2 \: \: }}$

$\mathtt{ \therefore \: \: x {}^{2} + y {}^{2} = (x + y) {}^{2} -2 xy} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ =36 - 2 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 34}$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: ANSWER \: \: } \mid}}}}}$

$\mathtt{ \: \: \star \: \: If \: \: \red{x = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1}} \: \: \:and \: \: \: \red{ y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1 }} \: ,\: then \: \: the \: \:} \\ \\ \mathtt{value \: \: of \: \: \: \: \: \underline{ \red { \: \: x {}^{2} + y {}^{2} = 34 \: \: }}.}$