Question

rationalize the denominator : 1/√2+√3

Answer 1

Given

$\frac{1}{ \sqrt{2} + \sqrt{3} }$
Rationalizing the denominator, we get

$\frac{1}{ \sqrt{2 } + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$
Now using a²-b²=(a+b) (a-b) in denominator we get
$\frac{ \sqrt{2} - \sqrt{3} }{ { (\sqrt{2}) }^{2} - ( { \sqrt{3} })^{2} }$
Square root cancelled with square

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

$\frac{ \sqrt{2} - \sqrt{3} }{ - 1}$

$- \sqrt{2} + \sqrt{3}$

$\sqrt{3} - \sqrt{2}$
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Answer 2

Step-by-step explanation:

Given:-

1/√2 + √3

To find out:-

Rationalising the denominator.

Solution:-

We have,

1/√2 + √3

Denominator is √2 + √3

we know that

Rationalising factor of √a + √b = √a - √b

so, rationalising factor of √2+√3 = √2-√3

On rationalising the denominator them

→ [1/(√2 + √3)]/[(√2 - √3)/(√2 - √3)]

→ [1(√2 - √3)]/[(√2 + √3)(√2 - √3)]

→ (√2 - √3)/[(√2 + √3)(√2 - √3)]

The denomination is in the form of (a+b)(a-b) = a^2-b^2

Where, we have to put in our equation a = √2 and b = √3 , we get

→ (√2 - √3)/[(√2)^2 - (√3)^2]

→ (√2 - √3)/(2 - 3)

→ (√2 - √3)/ -1

→ √3 - √2

Hence, the denominator is rationalised.

Answer:-

√3 - √2

Used formula:-

Rationalising factor of √a + √b = √a - √b

(a+b)(a-b) = a^2-b^2

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