Math, asked by shresthofficial65, 2 months ago


 \frac{1}{ \sqrt{7}  +  \sqrt{6} -  \sqrt{13}  }

Answers

Answered by Salmonpanna2022
8

Answer:

→ √6/12 + √7/14 + √546/84.

Step-by-step explanation:

Complete question:-

Given that:-

Rationalise the denominator of 5/√7+√6-√13.

What to do:-

To rationalise the denominator.

Solution:-

We have,

1/(√7+√6-√13)

Rationalising factor = √7+√6+√13

→ 1/(√7+√6-√13) × √7+√6+√13/(√7+√6+√13)

→ √7+√6+√13/(√7+√6-√13)(√7+√6+√13)

→ √7+√6+√13/(√7+√6)²-(√13)²

→ √7+√6+√13/(√7²+√6²+2(√7)(√6))-13

→ √7+√6+√13/13+2√42-13

→ √7+√6+√13/2√42

The denominator is still irrational. So we have to rationalise it further.

Again,

√7+√6+√13/2√42

Now rationalising factor = √42

→ √7+√6+√13/2√42 × √42/√42

→ √42(√7+√6+√13)/2(√42)²

→ √42×7+√42×6+√42×13/2(42)

→ (7√6+6√7+√546)/84

→ 7√6/84 + 6√7/84 + √546/84

→ √6/12 + √7/14 + √546/84 Ans.

Now the denominator is rationalised.

Hope it helps...☺

Answered by shreyaSingh2022
7

Step-by-step explanation:

1/(√7+√6-√13)

Rationalising factor = √7+√6+√13

→ 1/(√7+√6-√13) × √7+√6+√13/(√7+√6+√13)

→ √7+√6+√13/(√7+√6-√13)(√7+√6+√13)

→ √7+√6+√13/(√7+√6)²-(√13)²

→ √7+√6+√13/(√7²+√6²+2(√7)(√6))-13

→ √7+√6+√13/13+2√42-13

→ √7+√6+√13/2√42

The denominator is still irrational. So we have to rationalise it further.

Now rationalising factor = √42

→ √7+√6+√13/2√42 × √42/√42

→ √42(√7+√6+√13)/2(√42)²

→ √42×7+√42×6+√42×13/2(42)

→ (7√6+6√7+√546)/84

→ 7√6/84 + 6√7/84 + √546/84

→ √6/12 + √7/14 + √546/84

Now the denominator is rationalised.

Hope it helps.

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