Question

solution fast please​

Answer 1

Step-by-step explanation:

i hope it will helpful for you

Answer 2

Step-by-step explanation:

$\sqrt{3} - 1 \div \sqrt{3} + 1 \\ = ( \sqrt{3} - 1)( \sqrt{3} - 1) \div = ( \sqrt{3} + 1)( \sqrt{3} - 1) \\ = ( \sqrt{3 } - 1) {}^{2} \div ( { \sqrt{3} )}^{2} - {1}^{2} \\ = 3 + 1 - 2 \sqrt{3} \div 3 - 1 \\ = 4 - 2 \sqrt{3} \div 2 \\ = 2 - \sqrt{3} \\ a - b \sqrt{3} = 2 - \sqrt{3} \\ here \: a = 2 \: and \: b = 1$

$4 + \sqrt{2} \div 2 + \sqrt{2} \\ = (4 + \sqrt{2} )(2 - \sqrt{2} ) \div (2 + \sqrt{2} )(2 - \sqrt{2} ) \\ = 8 - 4 \sqrt{2} + 2 \sqrt{2} - 2 \div {2}^{2} - ( { \sqrt{2} )}^{2} \\ = 6 - 2\sqrt{2} \div 4 - 2 \\ 6 - 2 \sqrt{2} \div 2 \\ = 3 - \sqrt{2} \\ a - \sqrt{b} = 3 - \sqrt{2} \\ here \: a = 3 \: and \: b = 2$