Question

rationalise the denominator 6/2 under root 3 - under root 6
plz explain step by step​

Answer 1

Step-by-step explanation:

Rationalisingfactorof

( 3− 2)is

( 3+ 2 )

\begin{gathered}=\frac{\sqrt{6}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\\=\frac{\sqrt{6}(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2})}\\=\frac{\sqrt{6}(\sqrt{3}+\sqrt{2})}{3-2}\\=\sqrt{6}\times \sqrt{3} +\sqrt{6}\times \sqrt{2}\\=3\sqrt{2}+2\sqrt{3}\end{gathered}

=

( 3

−

2

)(

3

+

2

)

6

(

3

+

2

)

=

(

3

)

2

−(

2

)

2

)

6

(

3

+

2

)

=

3−2

6

(

3

+

2

)

=

6

×

3

+

6

×

2

=3

2

+2

3

Therefore,

\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}=3\sqrt{2}+2\sqrt{3}

3

−

2

6

=3

2

+2

3

Answer 2

Answer:

$\sqrt{3} -\sqrt{6}$

Step-by-step explanation:

$\frac{6}{2\sqrt{3} } - \sqrt{6}$

rationalizing means removing irrational number from denominator

so $\sqrt{3}$ needs to be removed

so if we multiply and divide with $\sqrt{3}$

then,

= $\frac{6*\sqrt{3} }{2\sqrt{3} *\sqrt{3} } - \sqrt{6}$

so,

= $\frac{6\sqrt{3} }{2*3 } - \sqrt{6}$

= $\frac{6\sqrt{3} }{6 } - \sqrt{6}$

= $\sqrt{3} -\sqrt{6}$

Hope it helps