Question

rationalise the denominators: 1) 3/√5+√2. 2) 2-√3/2+√3. 3)√3+1/√3-1. 4)√3-√2/√3+√2.

Answer 1

Here RFof the denominator = a binomial congugate to the expression

ie, (√2+√3)— (√5)

1/(√2+√3)+(√5) X (√2+√3)—(√5)/ (√2+√3)—(√5)

=(√2+√3—√5)/ (√2+√3)²—(√5)² (using identity (a+b)(a—b) = a² —b²)

= (√2+√3—√5) / 2+3+2√6

Answer 2

Hey there !!
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1) 3 / √5+√2

= 3 / (√5 + √2) × (√5 – √2) / (√5 – √2)

= 3 (√5 – √2) / [ (√5)² – (√2)²

= 3 (√5 – √2) / ( 5 – 2 )

= 3 (√5 – √2) / 3

= √5 – √2
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2) ( 2 – √3 ) / ( 2 + √3 )

= ( 2 – √3 ) / ( 2 + √3 ) × ( 2 – √3 ) / ( 2 – √3 )

= ( 2 – √3 )² / [ (2)² – (√3)² ]

= ( 4 + 3 – 4√3 ) / ( 4–3 )

= 7 – 4√3
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3) ( √3 – √2 ) / ( √3 + √2 )

= ( √3 – √2 ) / ( √3 + √2 ) × ( √3 – √2 ) / ( √3 – √2 )

= ( √3 – √2 )² / [ (√3)² – (√2)² ]

= ( 3 + 2 – 2√6 ) / ( 3 – 2 )

= 5 – 2√6
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Hope my ans.'s satisfactory. (^-^)