Math, asked by nithawani0814, 4 months ago

Rationalize the denominator 1/root 7+root 3-root2​

Answers

Answered by meenakshibhardwaj613
0

Step-by-step explanation:

Here, the given expression is,

\frac{1}{\sqrt{7} +\sqrt{3} -\sqrt{2} }

7

+

3

2

1

Multiply and divide by √7 + √3 + √2,

=\frac{1}{\sqrt{7} +\sqrt{3} -\sqrt{2} }\times \frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{\sqrt{7} +\sqrt{3} +\sqrt{2}}=

7

+

3

2

1

×

7

+

3

+

2

7

+

3

+

2

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

(

7

+

3

)

2

−(

2

)

2

7

+

3

+

2

=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{7+3+2\sqrt{21}-2}=

7+3+2

21

−2

7

+

3

+

2

=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{8+2\sqrt{21}}=

8+2

21

7

+

3

+

2

Multiply and divide by 8-2√21,

=\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{8+2\sqrt{21}}\times \frac{8-2\sqrt{21}}{8-2\sqrt{21}}=

8+2

21

7

+

3

+

2

×

8−2

21

8−2

21

=\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-2\sqrt{147} -2\sqrt{63} -2\sqrt{42}}{(8)^2-(2\sqrt{21})^2}=

(8)

2

−(2

21

)

2

8

7

+8

3

+9

2

−2

147

−2

63

−2

42

=\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-14\sqrt{3} -6\sqrt{7} -2\sqrt{42}}{64-84}=

64−84

8

7

+8

3

+9

2

−14

3

−6

7

−2

42

=\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-14\sqrt{3} -6\sqrt{7} -2\sqrt{42}}{-20}=

−20

8

7

+8

3

+9

2

−14

3

−6

7

−2

42

=\frac{2\sqrt{7} -6\sqrt{3} +9\sqrt{2}-2\sqrt{42}}{-20}=

−20

2

7

−6

3

+9

2

−2

42

=-\frac{1}{20}(2\sqrt{7}-6\sqrt{3}+8\sqrt{2}-2\sqrt{42})=−

20

1

(2

7

−6

3

+8

2

−2

42

)

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