Math, asked by unknown46r, 5 hours ago

9 / root 11 + root 5 rationalise​

Answers

Answered by adityagg973
0

Answer:

Answer:

\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}

6

+

5

11

1

=

60

6

5

+5

6

+

330

Given: \frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}

6

+

5

11

1

We have to rationalize the denominator.

Consider,

\frac{1}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}

(

6

+

5

)−

11

1

=\frac{1}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}\times\frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}=

(

6

+

5

)−

11

1

×

(

6

+

5

)−

11

(

6

+

5

)+

11

=\frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5}-\sqrt{11})(\sqrt{6}+\sqrt{5}+\sqrt{11})}=

(

6

+

5

11

)(

6

+

5

+

11

)

(

6

+

5

)+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}+\sqrt{5})^2-(\sqrt{11})^2}=

(

6

+

5

)

2

−(

11

)

2

6

+

5

+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6})^2+(\sqrt{5})^2+2\sqrt{6\times5}-11}=

(

6

)

2

+(

5

)

2

+2

6×5

−11

6

+

5

+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}=

2

30

6

+

5

+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}\times\frac{\sqrt{30}}{\sqrt{30}}=

2

30

6

+

5

+

11

×

30

30

=\frac{(\sqrt{6}+\sqrt{5}+\sqrt{11})\times\sqrt{30}}{2(\sqrt{30})^2}=

2(

30

)

2

(

6

+

5

+

11

30

=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{2\times30}=

2×30

6

5

+5

6

+

330

=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}=

60

6

5

+5

6

+

330

Answered by Krishrkpmlakv
1

Answer:

Step-by-step explanation:

Attachments:
Similar questions