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If x = 7 + 4√3 and xy = 1, then find the value of \(\frac{{{x^2}~ -~5xy ~+ \;{y^2}}}{{{x^2}~ + ~5xy ~+ \;{y^2}}}\)
1. 187/197
2. 241/671
3. 189/199
4. 199/189

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Correct Answer - Option 3 : 189/199

Given:

x = 7 + 4√3

xy = 1

Calculation:

xy = 1

⇒ y = 1/x

⇒ y = 1/(7 + 4√3) 

⇒ y = 7 - 4√3    

(x + y) = 7 + 4√3 + 7 - 4√3  

⇒ (x + y) = 14      ----(1)

(x - y) = 7 + 4√3 - 7 + 4√3

⇒ (x - y) = 8√3     ----(2)

We have to find the value of \(\frac{{{x^2}~ -~5xy ~+ \;{y^2}}}{{{x^2}~ + ~5xy ~+ \;{y^2}}}\)

⇒ \(\frac{{{x^2} ~- ~2xy ~+ \;{y^2} - ~3xy}}{{{x^2}~ +~ 2xy ~+ \;{y^2} ~+~ 3xy}}\)

⇒ \(\frac{{{{\left( {x\; - \;y} \right)}^2}\; - \;3xy}}{{{{\left( {x\; + \;y} \right)}^2}\; + \;3xy}}\)

From equation (1) and (2) put the value of (x + y) and (x- y) in the above expression, we get

⇒ {(8√3)2 - 3}/(142 + 3)

⇒ 189/199

∴ The value of \(\frac{{{x^2}~ -~5xy ~+ \;{y^2}}}{{{x^2}~ + ~5xy ~+ \;{y^2}}}\) is 189/199

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