Question

Rationalise the denominator of: 1/(3+√2 - 3√3)

Help!!​

Answer 1

Answer:

Solution

3

−

2

+1

1

=

(

3

−

2

)+1

1

×

(

3

−

2

)−1

(

3

−

2

)−1

=

(

3

−

2

)

2

−(1)

2

3

−

2

−1

=

(

3

)

2

−2

6

+(

2

)

2

−1

3

−

2

−1

=

3−2

6

+2−1

3

−

2

−1

=

4−2

6

3

−

2

−1

=

2(2−

6

)

(

3

−

2

)−1

=

2(2−

6

)

3

−

2

−1

×

2+

6

2+

6

=

2[(2)

2

−(

6

)

2

]

2

3

−2

2

−2+

18

−

12

−

6

=

2(4−6)

2

3

−2

2

−2+3

2

−2

3

−

6

=

2(−2)

2

−2−

6

=

−4

2

−2−

6

=

4

1

(2+

6

−

2

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Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

$\mathsf{\;\;Given :\;\dfrac{1}{3 + \sqrt{2} - 3\sqrt{3}}}$

$\mathsf{Multiplying\;numerator\;and\;denominator\;with\;3 + \sqrt{2} + 3\sqrt{3},\;We\;get :}$

$\mathsf{\implies \dfrac{3 + \sqrt{2} + 3\sqrt{3}}{(3 + \sqrt{2} - 3\sqrt{3})(3 + \sqrt{2} + 3\sqrt{3})}}$

★  We know that : (A + B)(A - B) = A² - B²

$\mathsf{\implies \dfrac{3 + \sqrt{2} + 3\sqrt{3}}{(3 + \sqrt{2})^2 - (3\sqrt{3})^2}}$

$\mathsf{\implies \dfrac{3 + \sqrt{2} + 3\sqrt{3}}{(3)^2 + (\sqrt{2})^2 + 2(3)\sqrt{2} - 9(\sqrt{3})^2}}$

$\mathsf{\implies \dfrac{3 + \sqrt{2} + 3\sqrt{3}}{9 + 2 + 6\sqrt{2} - 9(3)}}$

$\mathsf{\implies \dfrac{3 + \sqrt{2} + 3\sqrt{3}}{11 + 6\sqrt{2} - 27}}$

$\mathsf{\implies \dfrac{3 + \sqrt{2} + 3\sqrt{3}}{6\sqrt{2} - 16}}$

$\mathsf{Multiplying\;numerator\;and\;denominator\;with\;6\sqrt{2} + 16,\;We\;get :}$

$\mathsf{\implies \dfrac{(3 + \sqrt{2} + 3\sqrt{3})(6\sqrt{2} + 16)}{(6\sqrt{2} - 16)(6\sqrt{2} + 16)}}$

$\mathsf{\implies \dfrac{3(6\sqrt{2}) + 3(16) + 6\sqrt{2}(\sqrt{2}) + 16\sqrt{2} + 3\sqrt{3}(6\sqrt{2}) + 16(3\sqrt{3})}{(6\sqrt{2})^2 - (16)^2}}$

$\mathsf{\implies \dfrac{18\sqrt{2} + 48 + 6(2) + 16\sqrt{2} + 18\sqrt{6} + 48\sqrt{3}}{36(\sqrt{2})^2 - 256}}$

$\mathsf{\implies \dfrac{34\sqrt{2} + 48 + 12 + 18\sqrt{6} + 48\sqrt{3}}{36(2) - 256}}$

$\mathsf{\implies \dfrac{34\sqrt{2} + 60 + 18\sqrt{6} + 48\sqrt{3}}{72 - 256}}$

$\mathsf{\implies \dfrac{34\sqrt{2} + 60 + 18\sqrt{6} + 48\sqrt{3}}{-184}}$

$\mathsf{\implies \dfrac{2(17\sqrt{2} + 30 + 9\sqrt{6} + 24\sqrt{3})}{-184}}$

$\mathsf{\implies \dfrac{-17\sqrt{2} - 30 - 9\sqrt{6} - 24\sqrt{3}}{92}}$

$\bf \underline{Hence, the\: denominator\:is\: rationalised.}$