Question

rationalize the denominator of
1/7+3√3​

Answer 1

Answer:

1/7+3√3

for rationalising the denominator multiply the denominator and numerator with 7-3√3

1/7+3√3*7-3√3/7-3√3

7-3√3/(7)²-(3√3)²

7-3√3/49-27

7-3√3/22

Therefore denominator is rationalised

7-3√3/22

Answer 2

$\bf \red{\frac{1}{7 + 3 \sqrt{3} } = \frac{7 - 3 \sqrt{3} }{22} }$

Given:

• $\frac{1}{7 + 3 \sqrt{3} }$

To find:

• Rationalize the denominator.

Solution:

Concept to be used:

• Rationalization is a process to free the denominator from any radical sign.
• For that multiply both numerator and denominator by rationalization factor.
• Rationalization factor is conjugate of denominator.
• If denominator is $(a + \sqrt{b} )$ then RF is $(a - \sqrt{b} )$

Step 1:

Find RF of denominator.

As denominator is $7 + 3 \sqrt{3}$

Thus,

RF is $\bf 7 - 3 \sqrt{3}$

Step 2:

Multiply by RF.

$\frac{1}{7 + 3 \sqrt{3} } = \frac{1}{7 + 3 \sqrt{3} } \times \frac{7 - 3 \sqrt{3} }{7 - 3 \sqrt{3} }$

apply Identity $\bf (x - y)(x + y) = {x}^{2} - {y}^{2}$

or

$= \frac{7 - 3 \sqrt{3} }{( {7)}^{2} - ( {3 \sqrt{3}) }^{2} }$

or

$= \frac{7 - 3 \sqrt{3} }{49 - 27}$

or

$= \frac{7 - 3 \sqrt{3} }{22}$

Thus,

After rationalization, it becomes

$\bf \frac{1}{7 + 3 \sqrt{3} } = \frac{7 - 3 \sqrt{3} }{22}$

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