Question

pls solve the questions given in the above attachment

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Answer 1

Questions:-

1. Simplify the following by rationalising the denominator $\sf \dfrac{6 - 4\sqrt{3}}{6+4\sqrt{3}}$
2. Simplify $\sf (125)^{\frac{2}{3} } - \sqrt{25} \times 5^0 \times {\frac{1}{125} }^{\frac{-1}{2}}$

Answers:-

1)

Rationalising the denominator means multiplying the numerator and denominator by the conjugate of denominator i.e, the term in which the signs are changed.

So, we should multiply the given rational number by (6 - 4√3) which is the conjugate of (6 + 4√3).

$\implies \sf \: \frac{6 - 4 \sqrt{3} }{6 + 4 \sqrt{3} } \times \frac{(6 - 4 \sqrt{3} )}{(6 - 4 \sqrt{3}) } \\ \\ \\ \implies \sf \: \frac{ {(6 - 4 \sqrt{3}) }^{2} }{(6 + 4 \sqrt{3})(6 - 4 \sqrt{3} ) }$

Using (a - b)² = a² + b² - 2ab and (a + b)(a - b) = a² - b² we get,

$\implies \sf \: \frac{ {6}^{2} + {(4 \sqrt{3} )}^{2} - 2(6)(4 \sqrt{3} ) }{ {6}^{2} - {(4 \sqrt{ 3} })^{2} } \\ \\ \\ \implies \sf \: \frac{36 + 16 \times 3 - 48 \sqrt{3} }{36 - 16 \times 3} \\ \\ \\ \implies \sf \: \frac{84 - 48 \sqrt{3} }{36 - 48} \\ \\ \\ \implies \sf \: \frac{84 - 48 \sqrt{3} }{ - 12} \\ \\ \\ \implies \sf \: \frac{ - 12(4\sqrt{3} -7) }{ - 12} \\ \\ \\ \implies \boxed{ \sf \:4 \sqrt{3} - 7}$

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2) $\sf (125)^{\frac{2}{3} } - \sqrt{25} \times 5^0 \times {\frac{1}{125} }^{\frac{-1}{2}}$

Using 125 = 5³ & a⁰ = 1 & 1/a = a⁻¹ we get,

$\implies \sf \: {( {5}^{3} )}^{ \frac{2}{3} } - 5 \times 1 \times {( {125}^{ - 1} )}^{ \frac{ - 1}{2} }$

Using (aᵐ)ⁿ = aᵐⁿ we get,

$\implies \sf \: {5}^{ \frac{6}{3} } - 5 \times {(125)}^{ \frac{1}{2} }$

Using a^(1/n) = n√a we get,

$\implies \sf \: {5}^{2} - 5 \times \sqrt[2]{125} \\ \\ \\ \implies \sf \: 25 - 5 \sqrt{25 \times 5} \\ \\ \\ \implies \sf \: 25 - 5 \times 5 \sqrt{5} \\ \\ \\ \implies \boxed{ \sf \: 25 - 25 \sqrt{5} }$