Question

retionalize the denominator 5-2√3 /6+7√2​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$Rationalise the denominator 5-2√3 /6+7√2

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$\bold{= > \frac{5 - 2 \sqrt{3} }{6 + 7 \sqrt{2} }}$

• Now,we Rationalise it by multiplying with opposite signs of denominator to both numerator and denominator.

$\bold{= > \frac{5 - 2 \sqrt{3} }{6 + 7 \sqrt{2} } \times \frac{6 - 7 \sqrt{2} }{6 - 7 \sqrt{2} }}$

$\bold{= > \frac{5(6 - 7 \sqrt{2}) - 2 \sqrt{3}(6 - 7 \sqrt{2} )}{ {(6)}^{2} - {(7 \sqrt{2} )}^{2} }}$

$\bold{= > \frac{30 - 35 \sqrt{2} - 12 \sqrt{3} + 14 \sqrt{2} \sqrt{3} }{36 - 98} }$

$\bold{ = > \frac{30 - 35 \sqrt{2} - 12 \sqrt{3} + 14 \sqrt{2} \sqrt{3} }{ - 62} }$

$\bold{ = > \frac{ - (30 + 35 \sqrt{2} + 12 \sqrt{3} - 14 \sqrt{2} \sqrt{3}) }{ - 62}}$

$\bold{ = > \frac{30 + 35 \sqrt{2} + 12 \sqrt{3} - 14 \sqrt{2} \sqrt{3} }{62}}$

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