Math, asked by mannatnavlakha14, 7 months ago

Simplify 4 divide by
2+√3+√7
by rationalizing the denominator. ​

Answers

Answered by piyushchavi0047
0

Step-by-step explanation:

Answer:

\frac{4}{2+\sqrt{3}+\sqrt{7}}

2+

3

+

7

4

= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}

3

(2

3

+3−

21

)

Explanation:

Given \frac{4}{2+\sqrt{3}+\sqrt{7}}

2+

3

+

7

4

multiply numerator and denominator by (2+√3-√7), we get

=\frac{4(2+\sqrt{3}-\sqrt7)}{[(2+\sqrt{3})+\sqrt{7}][(2+\sqrt{3})-\sqrt{3}]}

[(2+

3

)+

7

][(2+

3

)−

3

]

4(2+

3

7

)

= \frac{4(2+\sqrt{3}-\sqrt{7})}{(2+\sqrt{3})^{2}-(\sqrt{7})^{2}}

(2+

3

)

2

−(

7

)

2

4(2+

3

7

)

= \frac{4(2+\sqrt{3}-\sqrt{7})}{2^{2}+2\times2\times\sqrt{3}+(\sqrt{3})^{2}-7}

2

2

+2×2×

3

+(

3

)

2

−7

4(2+

3

7

)

=\frac{4(2+\sqrt{3}-\sqrt{7})}{4+4\sqrt{3}+3-7}

4+4

3

+3−7

4(2+

3

7

)

=\frac{4(2+\sqrt{3}-\sqrt{7})}{4\sqrt{3}}

4

3

4(2+

3

7

)

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

3

(2+

3

7

)

Rationalising the denominator, we get

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

3

×

3

(2+

3

7

3

= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}

3

(2

3

+3−

21

)

Therefore,

\frac{4}{2+\sqrt{3}+\sqrt{7}}

2+

3

+

7

4

= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}

3

(2

3

+3−

21

)

Similar questions