Question

Solve it . rationalisation ch 1 of class 9

Answer 1

Answer:

B

$28 + 10 \sqrt{3} \\ = 25 + 3 + 10 \sqrt{3} \\ = {5}^{2} + { \sqrt{3} }^{2} + 2(5)( \sqrt{3}) \\ = (5 + \sqrt{3} ) {}^{2}$

Similarly

$97 + 56 \sqrt{3} = (4 \sqrt{3} + 7) {}^{2} \\ \\ 16 + 6 \sqrt{7} = (3 + \sqrt{7} ) {}^{2} \\ \\ 16 - 6 \sqrt{7} = (3 - \sqrt{7} ) {}^{2}$

Just pit these values and you are done

$5 + \sqrt{3} - (4 \sqrt{3} + 7) + \frac{ \sqrt{7} }{3 + \sqrt{7} - 3 + \sqrt{7} } \\ \\ = - \frac{3}{2} - 3 \sqrt{3} \\ \\ = \frac{ - 3 - 6 \sqrt{3} }{2}$