Math, asked by soabakther8899, 11 months ago

Express each one of the following with rational denominator:
(i) 1/ 3+√2 (ii)1/√6 - √5 (iii) 16/ √41 - 5
(iv)30/ 5√3 - 3√5 (v) 1/ 2√5 - √3 (vi)√3 + 1 / 2√2 - √3 (vii)6 - 4√2 / 6 + 4√2 (viii) 3√2 + 1 / 2√5 - 3 (ix) b² / √a² + b² + a

Answers

Answered by nikitasingh79
31

Given :  (i) 1/ 3 + √2  

The Rationalisation factor for 3 + √2 is  3 - √2 . Therefore , multiplying and dividing by 3 - √2 we have :  

= 1 × (3 - √2)/ (3 + √2) (3 - √2)

= (3 - √2)/ (3² - √2²)

[By using the identity : (a + b)(a - b) = a² – b²]

= (3 - √2)/ (9 - 2)

= (3 - √2)/7

 

 

(ii)1/√6 - √5  

The Rationalisation factor for √6 - √5 is  √6 + √5 . Therefore , multiplying and dividing by √6 + √5 we have :  

= 1× (√6 + √5 )/(√6 - √5) (√6 + √5)

= (√6 + √5 )/(√6² - √5²)

[By using the identity : (a + b)(a - b) = a² – b²]

= (√6 + √5 )/(6 - 5)

= (√6 + √5 )/1

= (√6 + √5 )

1/√6 - √5 = (√6 + √5 )

 

(iii) 16/ √41 - 5

The Rationalisation factor for √41 - 5 is  √41 + 5. Therefore , multiplying and dividing by √41 + 5 we have :  

= 16 × (√41 + 5)/ (√41 - 5)(√41 + 5)

= 16 (√41 + 5)/ (√41² - 5²)

= 16 (√41 + 5)/ (41 - 25)

= 16 (√41 + 5)/ 16

= √41 + 5

= 16/ √41 - 5 = √41 + 5

(iv) 30/ 5√3 - 3√5  

The Rationalisation factor for 5√3 - 3√5 is  5√3 + 3√5 . Therefore , multiplying and dividing by 5√3 + 3√5 we have :

= [30 × (5√3 + 3√5)] / (5√3 - 3√5)(5√3 + 3√5)

= [30 (5√3 + 3√5)] / [(5√3)² - (3√5)²]

= [30 (5√3 + 3√5)] / [(25 × 3) - (9 ×5)]

= [30 (5√3 + 3√5)] / [75 - 45]

= [30 (5√3 + 3√5)] / [30]

= (5√3 + 3√5)

30/ 5√3 - 3√5  =  (5√3 + 3√5)

 

(v) 1/ 2√5 - √3

The Rationalisation factor for 2√5 - √3 is  2√5 + √3. Therefore , multiplying and dividing by 2√5 + √3 we have :

= 1 × (2√5 + √3)/ (2√5 - √3)(2√5 + √3)

= (2√5 + √3)/ [(2√5)² - √3)²]

=  (2√5 + √3)/ [(4 × 5) - 3)]

=  (2√5 + √3)/ [20 - 3)]

=  [(2√5 + √3)]/ 17

1/ 2√5 - √3 = [(2√5 + √3)]/ 17

 

(vi) √3 + 1 / 2√2 - √3  

The Rationalisation factor for 2√2 - √3  is  2√2 + √3 . Therefore , multiplying and dividing by 2√2 + √3 we have :

= [(√3 + 1) × (2√2 + √3)] / [(2√2 - √3 )(2√2 + √3)]

= [√3 × 2√2 + √3 × √3 + 2√2 + √3]/ [(2√2)² - √3²]

= [2√6 + 3 + 2√2 + √3]/ [4 × 2 - 3]

= [2√6 + 3 + 2√2 + √3]/ [8 - 3]

= [2√6 + 3 + 2√2 + √3]/ 5

(vii) 6 - 4√2 / 6 + 4√2

The Rationalisation factor for 6 + 4√2 is  6 - 4√2. Therefore , multiplying and dividing by 6 - 4√2 we have :

= (6 - 4√2) × (6 - 4√2)  / (6 + 4√2)(6 - 4√2)

= (6 - 4√2)² /[6² -  (4√2)²]

= [6² + (4√2)² - 2 × 6 × 4√2]/ (36 - 16 × 2)

By Using Identity : (a - b)²  = a² + b² - 2ab & (a + b)(a – b) = a² - b²

= [36 + 16 × 2 - 48√2]/(36 - 32)

= [36 + 32 - 48√2] / 4

= (68 - 48√2) /4

= 4(17 - 12√2)/4

= 17 - 12√2

 

(viii) 3√2 + 1 / 2√5 - 3  

The Rationalisation factor for 2√5 - 3  is  2√5 + 3 . Therefore , multiplying and dividing by 2√5 + 3 we have :

= [(3√2 + 1) (2√5 + 3 )] / (2√5 - 3)(2√5 + 3)

= [3√2 × 2√5 + 3√2 × 3 + 2√5 × 1 + 3] /[(2√5)² - 3²]

= [6√10 + 9√2 + 2√5 + 3]/ [4 × 5 - 9]

= [6√10 + 9√2 + 2√5 + 3]/ [20 - 9]

= [6√10 + 9√2 + 2√5 + 3]/ 11

 

(ix) b² / √a² + b² + a

The Rationalisation factor for √a² + b² + a is  √a² + b² - a. Therefore , multiplying and dividing by √a² + b² - a we have :

= b² × (√a² + b² - a) / [(√a² + b² + a)(√a² + b² - a)]

= b² (√a² + b² - a) / [(√a² + b²)²  -  a²)]

= b² (√a² + b² - a) / [(a² + b² -  a²)]

= b² (√a² + b² - a) /  b²

= √a² + b² - a

 

HOPE THIS ANSWER WILL HELP YOU…..

 

Some questions of this chapter :

In each of the following determine rational numbers a and b.

(i) √3-1/√3+1= a-b√3 (ii)4+√2/2+√2= a-√b

(iii) 3 + √2 / 3-√2 = a+b √2

(iv) 5 + 3√3 / 7 + 4√3 = a+b√3 (v) √11 - √7 / √11 + √7 = a-b √77 (vi) 4 + 3√5 / 4 - 3√5 = a + b√5

https://brainly.in/question/15897356

Rationalies the denominator and simplify : (i) √3 - √2 / √3 + √2 (ii)5 + 2√3 / 7+ 4√3 (iii) 1+√2/ 3-2√2, (iv)2√6 - √5 / 3√5 - 2√6 (v)4√3 + 5√2 / √48 + √18, (vi)2√3 - √5 / 2√2 + 3√3

https://brainly.in/question/15897355


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