Question

what would be the denominator after rationalizing 7/(5root3 -5rot2)?
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Answer 1

7/(5√3-5√2)×(5√3+5√2)/(5√3+5√2)

35√3+35√2/75-50

35(√3+√2)/25

7(√3+√2)/5 answer

Answer 2

Answer:

$\frac{7}{5 \sqrt{3} - 5 \sqrt{2} } = \frac{7}{5 \sqrt{3} - 5 \sqrt{2} } \times \frac{5 \sqrt{3} + 5 \sqrt{2} }{5 \sqrt{3} + 5 \sqrt{2} } = \frac{7(5 \sqrt{3} + 5 \sqrt{2} )}{(5 \sqrt{3} - 5 \sqrt{2} )(5 \sqrt{3} + 5 \sqrt{2}) } = \frac{35( \sqrt{3} + \sqrt{2} )}{ {(5 \sqrt{3}) }^{2} - {(5 \sqrt{2}) }^{2} }$

$= \frac{35( \sqrt{3} + \sqrt{2} ) }{75 - 50} = \frac{35( \sqrt{3} + \sqrt{2} )}{25} = \frac{7 \sqrt{3} + 7 \sqrt{2} }{5}$

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