Question

7 under root 3 minus 5 under root 3 upon under root 48 + under root 48 Rationalise the denominator​

Answer 1

Step-by-step explanation:

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Here is the answer you were looking for:

Using the identity:

(x + y)(x - y) = {x}^{2} - {y}^{2}(x+y)(x−y)=x2−y2

\begin{gathered} \frac{7 \sqrt{3} - 5 \sqrt{2} }{ \sqrt{48} + \sqrt{18} } \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{ \sqrt{2 \times 2 \times 2 \times 2 \times 3} + \sqrt{3 \times 3 \times 2} } \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{ \sqrt{ {2}^{2} \times {2}^{2} \times 3 } + \sqrt{ {3}^{2} \times 2 } } \\ \\ = \frac{7 \sqrt{3}

- 5 \sqrt{2} }{2 \times 2 \sqrt{3} + 3 \sqrt{2} } \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \\ \end{gathered}48+1873−52=2×2×2×2×3+3×3×273−52=22×22×3+32×273−52=2×23+3273−52=43+3273−52

On rationalizing the denominator we get,

\begin{gathered} = \frac{7 \sqrt{3} - 5 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \times \frac{4 \sqrt{3} - 3 \sqrt{2} }{4 \sqrt{3} - 3 \sqrt{2} } \\ \\ = \frac{7 \sqrt{3} (4 \sqrt{3} - 3 \sqrt{2} ) - 5 \sqrt{2} (4 \sqrt{3} - 3 \sqrt{2}) }{ {(4 \sqrt{3}) }^{2} - {(3 \sqrt{2} )}^{2} } \\ \\ = \frac{84 - 21 \sqrt{6} - 20 \sqrt{6} + 30 }{48 - 18} \\ \\ = \frac{114 - 41 \sqrt{6} }{30} \end{gathered}=43+3273−52×43−3243−32=(43)2−(32)273(43−32)−52(43−32)=48−1884−216−206+30=30114−416

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