Question

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Answer 1

Hope u like my process
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Given
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$= > \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } = 1$

Now..

$= > a + b \sqrt{3} = \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } = 1 \\ \\ \: \: so \: \: comparing \: \: we \: get \: \: \\ \\ = > \boxed{ \bf \: a = \: \blue 1 \: \: and \: \: b = \blue0 }$
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Answer 2

✍✍HERE IS YOUR ANSWER..⬇⬇

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$\underline{SOLUTION}$ ➡

$Given ,\\ \\ \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3} \\ \\ = > a + b \sqrt{3} = \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = > a + b \sqrt{3} = 1$

$By \: \: comparing \: both \: sides ,\: \: we \: get,$

$\boxed{a = 1 \: \: \: and \: \: \: b = 0}$

$L.H.S.= \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ \: \: \: \: \: = 1$

$R.H.S. = a + b \sqrt{3} \\ \\ = 1 + 0 \times \sqrt{3} \: \: \: \: \: \: \: (by \: putting \: values \: of \: a \: \: and \: \: \: b)\\ \\ = 1 + 0 \\ \\ = 1$

$So, \\ L.H.S. \: = \: R.H.S.$
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