Math, asked by Dhyank, 9 months ago

rationalise the denominator of (i) (5/2+√2) (ii) (2√3+5)​

Answers

Answered by rksingh61553
0

Answer:

The given expression is

\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}

3

+

5

2

3

Rationalize the denominator.

\frac{3}{(\sqrt{3}+\sqrt{5})-\sqrt{2}}\times \frac{(\sqrt{3}+\sqrt{5})+\sqrt{2}}{(\sqrt{3}+\sqrt{5})+\sqrt{2}}

(

3

+

5

)−

2

3

×

(

3

+

5

)+

2

(

3

+

5

)+

2

\frac{3(\sqrt{3}+\sqrt{5})+3\sqrt{2}}{(\sqrt{3}+\sqrt{5})^2-(\sqrt{2})^2}

(

3

+

5

)

2

−(

2

)

2

3(

3

+

5

)+3

2

\frac{3\sqrt{3}+3\sqrt{5}+3\sqrt{2}}{3+5+2\sqrt{15}-2}

3+5+2

15

−2

3

3

+3

5

+3

2

\frac{3\sqrt{3}+3\sqrt{5}+3\sqrt{2}}{6+2\sqrt{15}}

6+2

15

3

3

+3

5

+3

2

Rationalize the denominator.

\frac{3(\sqrt{3}+\sqrt{5}+\sqrt{2})}{6+2\sqrt{15}}\times \frac{6-2\sqrt{15}}{6-2\sqrt{15}}

6+2

15

3(

3

+

5

+

2

)

×

6−2

15

6−2

15

\frac{3(\sqrt{3}+\sqrt{5}+\sqrt{2})(6-2\sqrt{15})}{6^2-(2\sqrt{15})^2}

6

2

−(2

15

)

2

3(

3

+

5

+

2

)(6−2

15

)

\frac{3(\sqrt{3}+\sqrt{5}+\sqrt{2})(6-2\sqrt{15})}{36-60}

36−60

3(

3

+

5

+

2

)(6−2

15

)

\frac{3(\sqrt{3}+\sqrt{5}+\sqrt{2})(6-2\sqrt{15})}{-24}

−24

3(

3

+

5

+

2

)(6−2

15

)

-\frac{(\sqrt{3}+\sqrt{5}+\sqrt{2})(6-2\sqrt{15})}{8}−

8

(

3

+

5

+

2

)(6−2

15

please mark in brainlist

)

\frac{\sqrt{3}}{2}-\frac{3\sqrt{2}}{4}+\frac{\sqrt{30}}{4}

2

3

4

3

2

+

4

30

Therefore the simplified form is \frac{\sqrt{3}}{2}-\frac{3\sqrt{2}}{4}+\frac{\sqrt{30}}{4}

2

3

4

3

2

+

4

30

Answered by sapna3325
0

Step-by-step explanation:

1)

5 \div 2 +  \sqrt{2}  \\  \\  \times  \: and \:  \div 2 -  \sqrt{2}  \\  \\

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