Question

Rationalize the denominator √6-√3/√2-√3

Answer 1

Step-by-step explanation:

=root 6 - root 3/root2-root3

=

$\sqrt{6} - \sqrt{3} \div \sqrt{2} - \sqrt{3 }$

$\sqrt{6} - \sqrt{3} \times \sqrt{2} + \sqrt{3} \div \sqrt{2} - \sqrt{3} \times \sqrt{2} + \sqrt{3}$

$\sqrt{6} - \sqrt{3} \times \sqrt{2} + \sqrt{3} \div \sqrt{2 {}^{2} } - \sqrt{3 {}^{2} }$

$\sqrt{6} - \sqrt{3 } \times \sqrt{2} + \sqrt{3} \div 4 - 9$

$\sqrt{6} - \sqrt{3} \times \sqrt{2} + \sqrt{3} \div - 5$

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Answer 2

Answer:

=root 6 - root 3/root2-root3

=

Step-by-step explanation: