Question

13. What would be the denominator after rationalizing 7/(5√3-5√2​

Answer 1

Answer:

7/(5√3-5√2)

=7(5√3+5√2)/(5√3-5√2)(5√3+5√2)

=7(5√3+5√2)/{(5√3)^2-(5√2)^2}

=7(5√3+5√2)/(75-50)

=7×5(√3+√2)/25

=7(√3+√2)/5

Answer 2

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$\longmapsto \sf \dfrac{7}{5 \sqrt{3} - 5\sqrt{2}}$

$\longmapsto \sf\dfrac{7}{5 \sqrt{3} - 5\sqrt{2}} \times \frac{5 \sqrt{3} + 5 \sqrt{2} }{5 \sqrt{3} + 5\sqrt{2}}$

$\longmapsto \sf \dfrac{35 \sqrt{3} + 35 \sqrt{2} }{ {(5 \sqrt{3}) }^{2} - {(5 \sqrt{2}) }^{2}}$

$\longmapsto \sf\dfrac{35( \sqrt{3} + \sqrt{2} )}{75 - 50}$

$\longmapsto \sf\dfrac{35( \sqrt{3} + \sqrt{2} )}{25}$

$\longmapsto \sf\dfrac{7( \sqrt{3} + \sqrt{2})}{5}$

∴ 5 is the denominator after rationalizing