Math, asked by arpitabiswas90, 4 months ago

help me please













?????​

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Answered by itsPapaKaHelicopter
10

Answer:

⇒ \frac{7 + 3 \sqrt{5} }{3 +  \sqrt{5} }  -  \frac{7 - 3 \sqrt{5} }{3 -  \sqrt{5} }  = a +  \sqrt{5b}

⇒ \frac{7 + 3 \sqrt{5} }{3 +  \sqrt{5} }  \times   \frac{3 -  \sqrt{5} }{3 -  \sqrt{5} }  -  \frac{7 - 3 \sqrt{5} }{3 -  \sqrt{5} }  \times  \frac{3 +  \sqrt{5} }{3 +  \sqrt{5} }  = a +  \sqrt{5b}

⇒ \frac{(7 + 3 \sqrt{5} )(3 -  \sqrt{5}) }{9 - 5}  -  \frac{(7 - 3 \sqrt{5})(3 +  \sqrt{5} ) }{9 - 5}  = a +  \sqrt{5b}

⇒ \frac{21 - 7 \sqrt{5}  + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5}   + 9 \sqrt{5} + 15 }{4}  = a +  \sqrt{5b}

⇒ \frac{ - 14 \sqrt{5}  + 18 \sqrt{5} }{4}  = a +  \sqrt{5b}

⇒ \frac{4 \sqrt{5} }{4}  = a +  \sqrt{5b}

⇒ \sqrt{5}  = a +  \sqrt{5b}

Hence:

 \textbf{Value of A = 0 and B = 1}

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Answered by sangameshsuntyan
4

Answer:

hope it may help

have a nice day

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