Question

rationalise the denominator of 1 ÷ root7 +root6-root12​

Answer 1

$\frac{1}{( \sqrt{7} + \sqrt{6}) - \sqrt{12} }$

Rationalising Factor of

$( \sqrt{7 } + \sqrt{6} ) - \sqrt{12 } = ( \sqrt{7} + \sqrt{6} ) + \sqrt{12}$

Multiply Both Numerator and Denominator with Rationalising Factor.

$\frac{1( \sqrt{7} + \sqrt{6}) + \sqrt{12} }{[( \sqrt{7} + \sqrt{6} ) - \sqrt{12}][( \sqrt{7} + \sqrt{6} ) + \sqrt{12} ]}$

We know a²-b²=(a+b)(a-b)

$\frac{ \sqrt{7} + \sqrt{6} + \sqrt{12} }{( \sqrt{7} + \sqrt{6}) {}^{2} - { \sqrt{12} }^{2} }$

We know (a+b)²=a²+b²+2ab

$\frac{ \sqrt{7} + \sqrt{6} + \sqrt{12} }{( { \sqrt{7} }^{2} + { \sqrt{6} }^{2} + 2 \sqrt{7 \times 6} ) - 12} \\ \\ \frac{ \sqrt{7} + \sqrt{6} + \sqrt{12} }{7 + 6 + 2 \sqrt{42} - 12} \\ \\ \frac{ \sqrt{7} + \sqrt{6} + \sqrt{12} }{13 - 12 + 2\sqrt{42} } \\ \\ \frac{ \sqrt{7} + \sqrt{6} + \sqrt{12} }{2 \sqrt{42} }$

√42 is Irrational Number Multiply Both Numerator and Denominator with √42

$\frac{( \sqrt{7} + \sqrt{6} + \sqrt{12}) \times \sqrt{42} }{2 \sqrt{42} \times \sqrt{42} } \\ \\ \frac{ \sqrt{7 \times 42} + \sqrt{6 \times 42} + \sqrt{42 \times 12} }{2 \times 42} \\ \\ \frac{ \sqrt{7 \times 7 \times 6} + \sqrt{6 \times 6 \times 7} + \sqrt{6 \times 7 \times 6 \times 2} }{84} \\ \\ \frac{7 \sqrt{6} + 6 \sqrt{7} + 6 \sqrt{14} }{84}$