Math, asked by parthk18, 1 year ago

if x=7+4√3 ,xy=1 then find 1/x square +1/y square

Answers

Answered by Anonymous
0
simple h ..1/x= y. ..
so .
1/xx=y² then .. u will get ams

Anonymous: kya hua tuje
Answered by TRISHNADEVI
4

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

 \underline{ \mathfrak{ \: Given, \: }} \\  \\ \:  \:  \:  \:  \:  \:  \tt{ x = 7 + 4 \sqrt{3} } \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{ xy = 1} \\  \\  \underline{ \mathfrak{ \:To \:  \:  find :-  \: }} \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt {\frac{1}{x {}^{2} }  +  \frac{1}{y  {}^{2}  }  =   \: ?}

 \underline{ \mathfrak{ \: Now, \: }} \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \tt{xy = 1 }\\ \\   \tt{\implies y =  \frac{1}{x} } \\  \\  \tt{\implies y =  \frac{1}{7 +4 \sqrt{3}  }}  \\  \\ \tt{  \implies y =  \frac{1(7  - 4 \sqrt{3} )}{(7 + 4 \sqrt{3})(7  - 4 \sqrt{3})  }}  \\  \\ \tt{\implies y  =  \frac{7 - 4 \sqrt{3} }{(7) {}^{2} - (4 \sqrt{3}  ) {}^{2} }}  \\  \\  \tt{\implies y  = \frac{7 - 4 \sqrt{3} }{49 - 48} } \\  \\  \tt{\implies y =  \frac{7 - 4 \sqrt{3} }{1}}  \\  \\   \:  \:  \:  \:  \:  \:  \tt{ \therefore \:  \: y= 7 - 4 \sqrt{3} }

 \underline{ \mathfrak{ \: Again, \: }} \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \sf{xy = 1} \\  \\  \sf{ \implies \: y =  \frac{1}{x} } \\  \\ \sf{ \implies \: y {}^{2}  = ( \frac{1}{x}) {}^{2}  } \\  \\ \sf{ \implies \: \frac{1}{x {}^{2} }  = y {}^{2} } \\  \\ \sf{ \implies \: \frac{1}{x {}^{2} }  = (7 - 4 \sqrt{3} ) {}^{2} } \\  \\  \sf{ \implies \: \frac{1}{x {}^{2} } = (7) {}^{2}   - 2 \times 7 \times  4\sqrt{3}  + (4 \sqrt{3} ) {}^{2} } \\  \\ \sf{ \implies \: \frac{1}{x {}^{2} } =49 - 56 \sqrt{3} + 48 } \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \sf{ \therefore \:  \:  \underline{ \:  \: \:  \frac{1}{x {}^{2} } =97 - 56 \sqrt{3} \:  \:  \:  }}

  \underline{\mathfrak{ \: And,  \: }} \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \sf{xy = 1} \\  \\  \sf{ \implies \: x =  \frac{1}{y} } \\  \\ \sf{ \implies \: x {}^{2}  = ( \frac{1}{y}) {}^{2}  } \\  \\  \sf{ \implies  \:  \frac{1}{y {}^{2} }  = x {}^{2} } \\  \\ \sf{ \implies \: \frac{1}{y {}^{2} }  = (7  + 4 \sqrt{3} ) {}^{2} } \\  \\  \sf{ \implies \: \frac{1}{y {}^{2} } = (7) {}^{2}    +  2 \times 7 \times  4\sqrt{3}  + (4 \sqrt{3} ) {}^{2} } \\  \\ \sf{ \implies \: \frac{1}{y {}^{2} } =49  +  56 \sqrt{3} + 48 } \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \sf{ \therefore \:  \:  \underline{ \:  \:  \: \frac{1}{y {}^{2} } =97  +  56 \sqrt{3} \:  \:  \:  \:  } \: }

 \bold{ \therefore \:  \:  \red{ \frac{1}{x {}^{2} }  +  \frac{1}{y {}^{2}}}  = (97 - 56 \sqrt{3} ) + (97 +  56\sqrt{3} )} \\  \\  \bold{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 97 - \cancel{ 56 \sqrt{3}} + 97 +  \cancel{56\sqrt{3}}} \\  \\  \bold{ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 97 + 97} \\  \\  \bold{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:=  \red{194}}

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