Question

Please ans my question

Answer 1

$\sf \: \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} } - \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } = a - b \sqrt{2}$

Rationalising the first term of L.H.S

For rationalising any term , we multiply the term with another term that has the same digits in numerator and denominator both but with the sign opposite to that of the denominator of first term .

$\sf \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} } = \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \times \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$

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

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

$= \frac{2 + 3 + 2 \sqrt{6} }{ - 1}$

$= \frac{5 + 2 \sqrt{6} }{ - 1}$

$= - 5 - 2 \sqrt{6}$

Similarly , Rationalising the second term

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

$= \frac{ \sqrt{2} {}^{2} - { \sqrt{3} }^{2} }{ { \sqrt{2} }^{2} - { \sqrt{3} }^{2} }$

$= \frac{ - 1}{ - 1}$

$= 1$

Solving the LHS

$\sf - 5 - 2 \sqrt{6} - 1 = a - b \sqrt{2}$

$\sf -6 -2 \sqrt{6} = a - b \sqrt{2}$

Comparing the coefficients ,

a = -6 and b = 2√3

Answer 2

Answer:

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