prove that
7 root 3 / root 10 + root 3 – 2 root 5 / root 6 + root 5 -3 root 2 / root 15 + 3 root 2 = 1​

1 thought on “prove that <br />7 root 3 / root 10 + root 3 – 2 root 5 / root 6 + root 5 -3 root 2 / root 15 + 3 root 2 = 1​”

1. Answer:

   Step-by-step explanation:

   7√3 / √10+√3 – 2√5 / √6 – 3√2 / √15 + 3√2

   Rationalise

   7√3 / √10+√3 = 7√3 ( √10 – √3 ) / ( √10+√3 ) ( √10-√3 )

   = 7√3 ( √10 – √3 ) / √10^2 – √3^2

   = 7√3 ( √10 – √3 ) / 10 – 3

   = 7√3 ( √10 – √3 ) /7

   = √3 ( √10 – √3 )

   = √30 – √9

   = √30 – 3

   2√5 / √6 + √5 = 2√5 ( √6 – √5 ) / (√6 + √5) ( √6 – √5 )

   = 2√30 – 10 / √6^2 – √5^2

   = 2√30 – 10 / 6-5

   = 2√30 – 10

   3√2 / √15 + 3√2 = 3√2 ( √15 – 3√2 ) / ( √15 + 3√2 ) ( √15 – 3√2 )

   = 3√2 ( √15 – 3√2 ) / √15^2 – 3√2^2

   = 3√2 ( √15 – 3√2 ) / 15 – 18

   = 3√2 ( √15 – 3√2 ) / -3

   = -√2 ( √15 – 3√2 )

   = -√30 + 6

   (7√3 / √10+√3) – (2√5 / √6 + √5 ) -( 3√2 / √15 + 3√2)

   = (√30 – 3 ) – ( 2√30 – 10 ) – ( -√30 + 6 )

   = √30 – 3 – 2√30 + 10 + √30 – 6

   = 1

   (7√3 / √10+√3) – (2√5 / √6 + √5 ) -( 3√2 / √15 + 3√2) = 1

   Hence proved

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