Math, asked by sushanthreddy8690, 9 months ago

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

Answers

Answered by mukulattri
0

Step-by-step explanation:

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Answered by nikitasingh79
1

(i) Given : (i) √3 - √2 / √3 + √2

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

= (√3 - √2) (√3 - √2) / (√3 + √2)(√3 - √2)

= (√3 - √2)²/√3² - √2²

= [√3² + √2² - 2 × √3 × √ 2] / 3 - 2

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

= [3 + 2 - 2√6]/1

= 5 - 2√6

(ii) Given : 5 + 2√3 / 7+ 4√3  

The Rationalisation factor for 7+ 4√3 is  7-  4√3 . Therefore , multiplying and dividing by 7 - 4√3 we have :  

= (5 + 2√3) × (7 - 4√3) / (7 + 4√3)(7 - 4√3 )

= [5 × 7 - 5 × 4√3 + 2√3 × 7 - 2√3 × 4√3[/ [(7² - (4√3)²]

= [35 - 20√3 + 14√3 - 8 × 3]/[49 - 16 × 3]

=  [35 - 6√3 + 14√3 - 24]/[49 - 48]

=  [35 - 24 + 6√3 ]/1

= 11 -  6√3  

 

(iii) Given : 1 + √2/ 3 - 2√2

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

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

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

= [3 + 2√2 + 3√2 + 2 × 2]/ [9 - 4 × 2]

= [3 + 5√2  + 4]/ [9 - 8]

= [3 + 4 + 5√2 ]/ 1

= 7 + 5√2

(iv) Given : 2√6 - √5 / 3√5 - 2√6  

The Rationalisation factor for 3√5 - 2√6  is  3√5 + 2√6 . Therefore , multiplying and dividing by 3√5 + 2√6  we have :

= [(2√6 - √5) × (3√5 + 2√6)] / [(3√5 - 2√6 )(3√5 + 2√6)]

= [2√6 × 3√5 + 2√6 × 2√6 - √5 × 3√5 - √5 × 2√6]/[(3√5)² - (2√6)²]

= [6√30 + 4 × 6 -  3 × 5 - 2√30]/[(9 × 5  - 4 × 6]

= [6√30 + 24 -  15 - 2√30]/[(45  - 24]

= [6√30  - 2√30 + 9]/[(21]

= (4√30 + 9)/21

(v) Given : 4√3 + 5√2 / √48 + √18

On simplying the denominator :  

√48 + √18 = √16 × 3 + √9 × 2

√48 + √18 = 4√3 + 3√2

The Rationalisation factor for 4√3 + 3√2  is  4√3 - 3√2 . Therefore , multiplying and dividing by 4√3 - 3√2 we have :

4√3 + 5√2 / 4√3 + 3√2  

= (4√3 + 5√2) × (4√3 - 3√2) / (4√3 + 3√2)(4√3 - 3√2)

= [4√3 × 4√3 - 4√3 × 3√2 + 5√2 × 4√3 - 5√2 × 3√2]/[(4√3)² - (3√2)²]

= [16 × 3 - 12√6 + 20√6 - 15 × 2]/ [16 × 3 - 9 × 2]

= [48 +  8√6 - 30]/[48 - 18]

= [48 - 30 + 8√6]/30

= (18 + 8√6)/30

= 2(9 + 4√6)/30

= (9 + 4√6)/15

(vi) Given : 2√3 - √5 / 2√2 + 3√3

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

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

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

= [4√6 - 6 × 3 - 2√10 + 3√15]/ [4 × 2 - 9 × 3]

= [4√6 - 18 - 2√10 + 3√15]/[8 - 27]

=  [4√6 - 18 - 2√10 + 3√15]/[- 19]

= (18 - 3√15 + 2√10 - 4√6]/19

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Some questions of this chapter :

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

https://brainly.in/question/15897352

Rationalise the denominator of each of the following (i-vii) :

(i) 3/√5 (ii)3/2√5 (iii)1/√12 (iv) √2/√5 (v) √3 + 1/√2 (vi) √2 + √5 / √3 (vii) 3√2/√5

https://brainly.in/question/15897353

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