Question

rationalize 2 root 3/root 6+2

Answer 1

Answer:

3

Step-by-step explanation:

Given, 33−936+23=a+b3

squaring on both sides, we get,

(a+b3)2 =33−1936+23×33+19333+193

=6312+1803=52+303

⇒a+b3=52+303

=52+215×15×3

=27+25

=3

Answer 2

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

$\huge\bold{:-2√3/√6+2}$

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$3

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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

$\huge\bold{ = > \frac{2 \sqrt{3} }{ \sqrt{6} + 2}}$

Here, To rationalise we multiply with opposite signs of denominator to both Numerator and denominator .

$\bold{ = > \frac{2 \sqrt{3} }{ \sqrt{6} + 2} \times \frac{ \sqrt{6} - 2 }{ \sqrt{6} - 2 }}$

$\bold{= > \frac{2 \sqrt{3}( \sqrt{6} - 2) }{ {( \sqrt{6} )}^{2} - {(2)}^{2} }}$

$\bold{ = > \frac{2 \sqrt{3} \sqrt{6} - 4 \sqrt{3} }{6 - 4} = \frac{2 \sqrt{3} \sqrt{6} - 4 \sqrt{3} }{2}}$

$\bold{= > \frac{2( \sqrt{3} \sqrt{6} - \sqrt{3}) }{2} = \sqrt{3} \sqrt{6} - \sqrt{3}}$

$\bold{\red{= > \sqrt{6} \sqrt{3} - \sqrt{3} = \sqrt{3} ( \sqrt{6} - 1)}}$

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