Question

plz ans this also....tomorrow is my exam

Answer 1

Hey user

Here is your answer :-

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$\sqrt{3 - 2 \sqrt{2} } \\ \\ let \: \sqrt{3 - 2 \sqrt{2} } = \sqrt{x} - \sqrt{y} \\ \\ squaring \: on \: the \: both \: sides \: \\ \\ ({ \sqrt{3 - 2 \sqrt{2} } )}^{2} = { (\sqrt{x} - \sqrt{y} )}^{2} \\ \\ 3 - 2 \sqrt{2} = x + y - 2 \sqrt{xy} \\ \\ x + y = 3 \\ \\ - 2 \sqrt{xy} = - 2 \sqrt{2} \\ \\ \sqrt{xy} = \sqrt{2} \\ \\ xy = 2 \\ \\ x = 2 \: \: \: \: \: and \: \: \: \: \: \: y = 1 \\ \\ hence \\ \\ \sqrt{3 - 2 \sqrt{2} } = \sqrt{2} - \sqrt{1} \\ \\ = \sqrt{2}$ - 1

Answer 2

$\sqrt{2 + 1 - 2 \sqrt{2} }$
$\sqrt{( \sqrt{2}) { }^{2} - 2 \sqrt{2} + (1) ^{2} }$
$\sqrt{( \sqrt{2} - 1) {}^{2} }$
therefore;
$\sqrt{2} - 1$
using formulae
$(a - b) ^{2}$
both answer is correct: