Question

5+2root3÷7+4root3×7-4root3÷7-4root3=m+nroot3

Answer 1

Hello this is your answer.. but i can do only till here....

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\bf \underline{What\: to\: do-}$

To rationalise the denominator

$\bf \underline{Solution-}$

$\textsf{We have,}$

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }$

$\textsf{The denominator is 5+2√3. Multiplying the numerator and denominator by 7-4√3,}$

$\textsf{we get,}$

$⟹\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

$⟹ \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 + 4 \sqrt{3} )(7 - 4 \sqrt{3}) }$

$\textsf{⬤ Applying Algebraic Identity</p><p>(a+b)(a-b) = a² - b² to the denominator}$

$\textsf{We get,}$

$⟹ \frac{(5 + 2 \sqrt{3} )(7 - 4 \sqrt{3}) }{(7 {)}^{2} - (4 \sqrt{3} {)}^{2} }$

$⟹ \frac{35 + 14 \sqrt{3} - 20 \sqrt{3} - 8 \sqrt{3 \times 3} }{49 - 48}$

$⟹ \frac{35 + 14 \sqrt{3} - 20 \sqrt{3} - 24 }{1}$

$⟹(35 - 24) - 6 \sqrt{3}$

$⟹11 - 6 \sqrt{3} \: \: \: \tt \red{ Ans}.$

$\bf \underline{Hence \:the \:denominator\: is\: rationalised.}$