Question

find the value a and b if √2+1÷√2-1 -√2-1÷√2+1=a+√2b

Answer 1

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

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

$\implies \frac{ 4 \times \sqrt{2} \times 1 }{2-1} = a +b \sqrt{2}$

$Here \:we \:used \: following \: algebraic \\identities$

1. (a+b)² - (a-b)² = 4ab
2. (a+b)(a-b) = a² - b²

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

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

/* Compare bothsides, we get */

$a = 0 \: and \: b = 4$

Therefore.,

$\green { a = 0 \: and \: b = 4 }$

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Answer 2

Answer:

a=0, b=4

i hope this helps