rationalise
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Q. Rationalize 1 / ( √5 + √6 - √11 ).
Solution :
= 1 ÷ ( √5 + √6 - √11 )
= 1 ÷ [ ( √5 + √6 ) - √11 ]
By multiplying numerator and denominator by [ ( √5 + √6 ) + √11 ].
= 1 [ ( √5 + √6 ) + √11 ] ÷ [ ( √5 + √6 ) - √11 ] [ ( √5 + √6 ) + √11 ]
= [ ( √5 + √6 ) + √11 ] ÷ [ ( √5 + √6 )² - ( √11 )² ]
= [ √5 + √6 + √11 ] ÷ [ (√5)² + (√6)² + 2 × √5 × √6 - (√11)² ]
= ( √5 + √6 + √11 ) ÷ ( 5 + 6 - 11 + 2√30 )
= ( √5 + √6 + √11 )÷ ( 2√30 )
By multiplying numerator and denominator by √30.
= √30 ( √5 + √6 + √11 ) ÷ ( 2√30 )√30
= ( √150 + √180 + √330 ) ÷ ( 2 × 30 )
=[( √(2 × 3 × 5 × 5 ) + √( 2 × 2 × 3 × 3 × 5 ) + √330 ] ÷ 60
= ( 5√6 + 6√5 + √330 ) / 60.
Solution :
= 1 ÷ ( √5 + √6 - √11 )
= 1 ÷ [ ( √5 + √6 ) - √11 ]
By multiplying numerator and denominator by [ ( √5 + √6 ) + √11 ].
= 1 [ ( √5 + √6 ) + √11 ] ÷ [ ( √5 + √6 ) - √11 ] [ ( √5 + √6 ) + √11 ]
= [ ( √5 + √6 ) + √11 ] ÷ [ ( √5 + √6 )² - ( √11 )² ]
= [ √5 + √6 + √11 ] ÷ [ (√5)² + (√6)² + 2 × √5 × √6 - (√11)² ]
= ( √5 + √6 + √11 ) ÷ ( 5 + 6 - 11 + 2√30 )
= ( √5 + √6 + √11 )÷ ( 2√30 )
By multiplying numerator and denominator by √30.
= √30 ( √5 + √6 + √11 ) ÷ ( 2√30 )√30
= ( √150 + √180 + √330 ) ÷ ( 2 × 30 )
=[( √(2 × 3 × 5 × 5 ) + √( 2 × 2 × 3 × 3 × 5 ) + √330 ] ÷ 60
= ( 5√6 + 6√5 + √330 ) / 60.
Anonymous:
Thanks Shubham
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