Question

Rationlise the denominator 3/9-3√2​

Answer 1

Answer:

The rationalised denominator of the given fraction is

$\displaystyle\boxed{\red{\sf\:\dfrac{3\:+\:\sqrt{2}}{7}}}$

Step-by-step-explanation:

We have to rationalise the given fraction.

$\displaystyle\sf\:\dfrac{3}{9\:-\:3\:\sqrt{2}}$

Now,

$\displaystyle\sf\:\dfrac{3}{9\:-\:3\:\sqrt{2}}$

$\displaystyle\implies\sf\:\dfrac{3}{9\:-\:3\:\sqrt{2}}\:\times\:\dfrac{9\:+\:3\:\sqrt{2}}{9\:+\:3\:\sqrt{2}}$

$\displaystyle{\implies\sf\:\dfrac{27\:+\:9\:\sqrt{2}}{\:(\:9\:)^2\:-\:(\:3\:\sqrt{2}\:)^2}\:\:\:-\:-\:[\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:=\:a^2\:-\:b^2\:]}$

$\displaystyle{\implies\sf\:\dfrac{27\:+\:9\:\sqrt{2}}{9\:\times\:9\:-\:3\:\times\:3\:\times\:\sqrt{2}\:\times\:\sqrt{2}}}$

$\displaystyle\implies\sf\:\dfrac{27\:+\:9\:\sqrt{2}}{81\:-\:9\:\times\:2}$

$\displaystyle\implies\sf\:\dfrac{27\:+\:9\:\sqrt{2}}{81\:-\:18}$

$\displaystyle\implies\sf\:\dfrac{27\:+\:9\:\sqrt{2}}{63}$

$\displaystyle\implies\sf\:\dfrac{\cancel{9}\:(\:3\:+\:\sqrt{2}\:)}{\cancel{63}}$

$\displaystyle\implies\boxed{\red{\sf\:\dfrac{3\:+\:\sqrt{2}}{7}}}$

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Additional Information:

1. Rationalised denominator:

When the denominator of a fraction is a rational number, then the denominator of the fraction is called rationalised denominator.

2. Rationalisation of denominator:

To rationalise the denominator of a fraction, we have to multiply both numerator and denominator by the conjugate pair of denominator.

3. Conjugate pair:

If the given number is sum of two numbers, then its conjugate pair is subtraction of those two numbers.

For example,

The conjugate pair of $\displaystyle\boxed{\red{\sf{4\:+\:\sqrt{5}}}}$ is $\displaystyle\boxed{\red{\sf{4\:-\:\sqrt{5}}}}$.

Answer 2

Answer:

★ ANSWER :

The rationalised form of $\dfrac{3}{9 \: - \: 3 \sqrt{2} }$ is $\dfrac{3 \: + \: \sqrt{2} }{7}$.

Step-by-step explanation:

★ QUESTION :

Rationlise the denominator $\dfrac{3}{9 \: - \: 3 \sqrt{2} }$.

★ CONCEPT USED :

• (a + b) (a - b) = a² - b²
• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab

★ SOLUTION :

$\frac{3}{9 \: - \: 3 \sqrt{2} } \\ \\ = \: \frac{3}{9 \: - \: 3 \sqrt{2} } \: \times \: \frac{9 \: + \: 3 \sqrt{2} }{9 \: + \: 3 \sqrt{2} } \\ \\ = \: \frac{3(9 + 3 \sqrt{2}) }{(9 \: - \: 3 \sqrt{2} ) \: (9 \: + \: 3 \sqrt{2}) } \\ \\ = \: \frac{27 \: + \: 9 \sqrt{2} }{ ({9}^{2} ) \: - \: (3 \sqrt{2})^{2} } \\ \\ = \: \frac{27 \: + \: 9 \sqrt{2} }{81 \: - (3 \times 3 \times \: \sqrt{2} \times \sqrt{2} ) } \\ \\ = \frac{27 \: + \: 9 \sqrt{2} }{81 \: - \: (9 \: \times \: 2)} \\ \\ = \frac{27 \: + \: 9 \sqrt{2} }{81 \: - \: 18} \\ \\ = \frac{27 \: + \: 9 \sqrt{2} }{63} \\ \\ = \frac{9(3 \: + \: \sqrt{2}) }{9 \: \times \: 7} \\ \\ = \: \frac{ \not \cancel9(3 \: + \: \sqrt{2}) }{ \not \cancel9 \: \times \: 7} \\ \\ = \: \frac{3 \: + \: \sqrt{2} }{7}$

★ ANSWER :

The rationalised form of $\dfrac{3}{9 \: - \: 3 \sqrt{2} }$ is $\dfrac{3 \: + \: \sqrt{2} }{7}$.

HOPE IT HELPS YOU !

THANKS !