Question

answer my questions and I'll give you thanks and vote you and Mark as brainalist if the answer is correct it's an urgent​

Answer 1

Answer:

4√2=a+√2b so by comparing coefficients a=0 and b=4

3.35353535...this will continue

so it is approximately

3.354 upto 3 places of decimal

Answer 2

Question no 2 is in the attachment

2)Given -

$\frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } - \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } = a + \sqrt{2} b$

To find -

The values of a and b

Solution -

$\frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } - \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } = a + \sqrt{2} b$

By rationalising the denominator,

$\frac{( \sqrt{2} + 1)( \sqrt{2} + 1)}{( \sqrt{2} - 1)( \sqrt{2} + 1) } - \frac{( \sqrt{2} - 1)( \sqrt{2} - 1)}{( \sqrt{2} + 1)( \sqrt{2} - 1) } = a + \sqrt{2} b$

$\frac{ {( \sqrt{2} + 1) }^{2} }{ {( \sqrt{2}) }^{2} - {(1)}^{2} } - \frac{ {( \sqrt{2} - 1)}^{2} }{ {( \sqrt{2}) }^{2} - {(1)}^{2} } = a + \sqrt{2} b$

$\frac{ {( \sqrt{2} )}^{2} + {(1)}^{2} + 2 \times \sqrt{2} \times 1}{2 - 1} - \frac{ {( \sqrt{2} )}^{2} + {(1)}^{2} - 2 \times \sqrt{2} \times 1 }{2 - 1} = a + \sqrt{2} b$

$\frac{2 + 1 + 2 \sqrt{2} }{1} - \frac{2 + 1 - 2 \sqrt{2} }{1} = a + \sqrt{2} b$

$3 + 2 \sqrt{2} - (3 - 2 \sqrt{2} ) = a + \sqrt{2} b$

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

$0 + 4 \sqrt{2} = a + \sqrt{2} b$

We know that rational numbers is always equal to other rational numbers and irrational numbers is always equal to other irrational numbers

$a = 0$

And,

$4 \sqrt{2} = \sqrt{2} b$

$b = 4$

Therefore the value of a is 0 and b is 4