Question

√3+√2/√3-√2 rationalise Tue denominator with steps ​

Answer 1

Step-by-step explanation:

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

multiple by

$\sqrt{3} + \sqrt{2}$

so we get

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

which is equal to

Answer 2

Answer:

5+2√6

Step-by-step explanation:

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

Multiply both denominator and Numerator by √3 +√2

Now open the identity: (a + b) ² in the numerator and (a²-b²) in the denominator

$\frac{ \sqrt{3}^{2} + \sqrt{2 }^{2} + (2 \times \sqrt{3} \times \sqrt{2}) }{3 - 2} = \frac{3 + 2 + 2 \sqrt{6} }{1} = 5 + 2\sqrt{6}$