Math, asked by mikasaaa, 3 months ago

simplify (Class 9th) ​

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Answers

Answered by ananta5088
1

denominator of each term.

\implies\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}\times\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{3}}\times\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}⟹

2

+

3

2

6

×

2

3

2

3

+

6

+

3

6

2

×

6

3

6

3

6

+

3

8

3

×

6

3

6

3

\implies\frac{4\sqrt{3}-6\sqrt{2}}{2-3}+\frac{12\sqrt{3}-6\sqrt{6}}{6-3}-\frac{24\sqrt{2}-24}{6-3}⟹

2−3

4

3

−6

2

+

6−3

12

3

−6

6

6−3

24

2

−24

\implies\frac{4\sqrt{3}-6\sqrt{2}}{-1}+\frac{12\sqrt{3}-6\sqrt{6}}{3}-\frac{24\sqrt{2}-24}{3}⟹

−1

4

3

−6

2

+

3

12

3

−6

6

3

24

2

−24

\implies-4\sqrt{3}+6\sqrt{2}+4\sqrt{3}-2\sqrt{6}-8\sqrt{2}+8⟹−4

3

+6

2

+4

3

−2

6

−8

2

+8

\implies8-2\sqrt{6}-2\sqrt{2}⟹8−2

6

−2

2

Therefore, \frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{3}}=8-2\sqrt{6}-2\sqrt{2}

2

+

3

2

6

+

6

+

3

6

2

6

+

3

8

3

=8−2

6

−2

2

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