Math, asked by maraheem724, 6 months ago

5. Rationalise the denominators of the following
1/3+√2, 1/√7-√6 ,1/√7,√6/√3-√2
plzzz answer if you know only​

Answers

Answered by pari2008chitra61
0

Step-by-step explanation:

Answer:

\frac{1}{\sqrt{7}- \sqrt{6}}

7

6

1

= (\sqrt{7}+\sqrt{6})(

7

+

6

)

Step-by-step explanation:

Given \frac{1}{\sqrt{7}- \sqrt{6}}

7

6

1

Multiply numerator and denominator by (√7+√6), we get

= \frac{(\sqrt{7}+\sqrt{6})} {(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}

(

7

6

)(

7

+

6

)

(

7

+

6

)

\* By algebraic identity:

(a+b)(a-b) = a²-b²

= \frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}

(

7

)

2

−(

6

)

2

(

7

+

6

)

= \frac{(\sqrt{7}+\sqrt{6})}{7-6}

7−6

(

7

+

6

)

= (\sqrt{7}+\sqrt{6})(

7

+

6

)

Therefore,

\frac{1}{\sqrt{7}- \sqrt{6}}

7

6

1

= (\sqrt{7}+\sqrt{6})(

7

+

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