prove that( 1/2+√3)+(1/√5-3)+(1+2-√5)=0
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Given, 1/(2 + √3) + 2/(√5 - √3) + 1/(2 - √5)
After rationalizing each term, we get
= [{1*(2 - √3)}/{(2 + √3)*(2 - √3)}] + [{2*(√5 + √3)}/{(√5 - √3)*(√5 + √3)}] + [{1*(2 + √5)}/{(2 + √5)*(2 - √5)}]
= (2 - √3)/(4 - 3) + {2*(√5 + √3)}/(5 - 3) + (2 + √5)/(4 - 5)
= (2 - √3) + {2*(√5 + √3)}/2 + (2 + √5)/(-1)
= (2 - √3) + (√5 + √3) - (2 + √5)
= 2 - √3 + √5 + √3 - 2 - √5
= 0
So, 1/(2 + √3) + 2/(√5 - √3) + 1/(2 - √5) = 0
After rationalizing each term, we get
= [{1*(2 - √3)}/{(2 + √3)*(2 - √3)}] + [{2*(√5 + √3)}/{(√5 - √3)*(√5 + √3)}] + [{1*(2 + √5)}/{(2 + √5)*(2 - √5)}]
= (2 - √3)/(4 - 3) + {2*(√5 + √3)}/(5 - 3) + (2 + √5)/(4 - 5)
= (2 - √3) + {2*(√5 + √3)}/2 + (2 + √5)/(-1)
= (2 - √3) + (√5 + √3) - (2 + √5)
= 2 - √3 + √5 + √3 - 2 - √5
= 0
So, 1/(2 + √3) + 2/(√5 - √3) + 1/(2 - √5) = 0
soddi:
thanks a lot ..i was confused
you have copied wrong question,... answer is correct
ya because like this type of exm..is little difficult to type
ok fine
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