Question

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Answer 1

Answer:

1/11

Step-by-step explanation:

2square =4

3square =6

1/1+4+6

1/11

I AM NOT SURE ABOUT THE ANSWER

SORRY

Answer 2

Question:

• Rationalise the denominator of $\dfrac{1}{1+\sqrt2 + \sqrt3 }$

Solution:

$\blue\underbrace{\red{\textbf{Points to know }}}:$

• Rationalising is method of removing irrational number or radial from the denominator.
• To Rationalise a fraction we have to multiply and divide the given fraction by the conjugate of the denominator.

$\longmapsto\dfrac{1}{(1+\sqrt2) + \sqrt3 }$

Here we have denominator ( 1 +√2 )+√3 . So, Its Conjugate will be  ( 1 +√2 ) -√3. So, We will multiply and divide the fraction by ( 1 +√2 ) -√3 .

$\longmapsto\dfrac{1}{(1+\sqrt2 )+ \sqrt3 } \\\\\\\dashrightarrow \dfrac{1}{(1+\sqrt2 )+ \sqrt3 } \times \dfrac{(1+\sqrt2 )- \sqrt3}{(1+\sqrt2 )- \sqrt3} \\\\\\\dashrightarrow \dfrac{(1+\sqrt2 )- \sqrt3}{\{(1+\sqrt2 )+\sqrt3\}\{(1+\sqrt2 )- \sqrt3\}}\\\\\\\dashrightarrow \dfrac{1+\sqrt2 - \sqrt3}{(1+\sqrt2 )^2- (\sqrt3)^2}\\\\\\\dashrightarrow \dfrac{1+\sqrt2 - \sqrt3}{(1^2+\sqrt2^2+\sqrt2\times 1\times2) - (3)}\\\\\\\dashrightarrow \dfrac{1+\sqrt2 - \sqrt3}{(1+2+2\sqrt2)-(3)}$

$\\\dashrightarrow \dfrac{1+\sqrt2-\sqrt3}{3+2\sqrt2-3}\\\\\\\dashrightarrow \dfrac{1+\sqrt2-\sqrt3}{2\sqrt2}$

• Here we have to again rationalise the fraction because there is still a irrational number in denominator.

$\dashrightarrow \dfrac{1+\sqrt2-\sqrt3}{2\sqrt2}\\\\\\\dashrightarrow \dfrac{1+\sqrt2-\sqrt3}{2\sqrt2}\times \dfrac{2\sqrt2}{2\sqrt2}\\\\\\\dashrightarrow \dfrac{2\sqrt2(1+\sqrt2-\sqrt3)}{8}\\\\\\\dashrightarrow \dfrac{\sqrt2(1+\sqrt2-\sqrt3)}{4}$

Hence,

• $\dfrac{1}{1+\sqrt2 + \sqrt3 }=\dfrac{\sqrt2(1+\sqrt2-\sqrt3)}{4}$