Question

solve these question pls..
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Answer 1

$\: \: \: \: \: \: \: \: \huge\colorbox{pink}{\tt{Answer ࿐}} \\ \\ \\ \\ {\sf{Let's \: assume \: that \: 5 \sqrt{5} + 2 \: is \: rational \: no. }} \\ \\ {\sf{\pink{∴ \: 5 \sqrt{5} + 2 \: = \frac{p}{q} \: ─━⧼ \: p \: and \: q \: are \: integers, \: q≠0}}} \\ \\ {\sf{∴ \: 5 \sqrt{5} + 2 = \frac{a}{b} \: ─━⧼ \: a \: and \: b \: are \: co-primes }} \\ \\ {\sf{⇝5 \sqrt{5} = \frac{a}{b} - 2 }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ {\sf{⇝ \sqrt{5} = \frac{a - 2b}{5b} }} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ {\sf{\color{blue}{∴ √5 = \frac{a - 2b}{5b}}}} \\ \\ {\sf{\green{∴ \sqrt{5} \: is \: a \: rational \: no. \: \: }}} \\ \\ {\sf{But, \: as \: we \: know \: that, \: \sqrt{5 } \: is \: irrational \: no. }} \\ \\ {\sf{∴This \: is \: contradiction. \: \: \: \: \: }} \\ \\ {\sf{Hence, \: our \: assumption \: is \: wrong.}} \\ \\ {\underline{\underline{\sf{\color{purple}{∴ \: 5 \sqrt{5} + 2 \: is \: an \: irrational \: no. }}}}} \\ \\ {\sf{\pink{Hence, \: proved.}}}$

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Answer 2

Answer:

Ok bro

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On 1:30