Question

pls send the 22nd question

Answer 1

a=16,b=-1
Hope it helps you
plz mark as brainliest

Answer 2

Heres your answer

Lets take the first rationalise question

$\frac{2 + \sqrt{3} }{2 - \sqrt{3} } ....x \\ \\ \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } = \frac{4+ 3 + 4 \sqrt{3} }{4- 3} = \frac{7 + 4 \sqrt{3} }{ 1} \\ = 7 + 4 \sqrt{3}$

Then in Second case
$\frac{2 - \sqrt{3} }{2 + \sqrt{3} } ....y \\ \\ \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 + \sqrt{3} } = \frac{4 + 3 - 4 \sqrt{3} }{4 - 3} = 7 - 4 \sqrt{3}$
And in 3 rd case

$\frac{ \sqrt{ 3} - 1 }{ \sqrt{3 + 1} } .....z \\ \\ \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 } = \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} = \frac{4 - 2 \sqrt{3} }{2} \\ \\ = \frac{2(2 - \sqrt{3} )}{2} = 2 - \sqrt{3}$

as
$x + y + z = a + b \sqrt{3} \\ \\ 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} + 2 - \sqrt{3} = a + b \sqrt{3} \\ \\ 7 + 7 + 2 - \sqrt{3} = a + b \sqrt{3} \\ \\ 16 - \sqrt{3} = a + b \sqrt{3} \\ \\ so \: a = 16 \: and \: b = - 1$

Hope it Helps!!