Question

1/2+root3 +2/root5-root3 +1/2-root5 =0​

Answer 1

Answer:

Hi ,

a ) 1/ ( 2 + √3 )

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

= ( 2 - √3 ) / [ 4 - 3 ]

= 2 - √3 ---( 1 )

b ) 2 / ( √5 - √3 )

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

= 2( √5 + √3 ) / ( 5 - 3 )

= 2 ( √5 + √3 )/ 2

= √5 + √3 ----( 2 )

c ) 1/( 2 - √5 )

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

= ( 2 + √5 ) / [ 4 - 5 ]

√= - ( 2 + √5 ) ---- ( 3 )

according to the problem given ,

( 1 ) + ( 2 ) + ( 3 )

= 2 - √3 + √5 + √3 - ( 2 + √5 )

= 2 - √3 + √5 + √3 - 2 - √5

= 0

I hope this helps you.

Answer 2

Step-by-step explanation:

Given question :

1 2 1

____ + _____ + _____ = 0

2+√3 √5-√3 2-√5

Answer :

First we have to rationalize the denominator.

1 1 2-√3 2-√3 2-√3

____ = ____ × _____ = _____ = _____

2+√3 2+√3 2-√3 2²-√3² 4-3

= 2-√3

2 2 √5+√3 2√5+2√3

_____ = _____ × ______ = _________

√5-√3 √5-√3 √5+√3 5 - 3

= 2√5+2√3 / 2 = √5+√3

1 1 2+√5 2+√5 - (2+√5)

____ = ____ × ____ = _____ = ______

2-√5 2-√5 2+√5 4 - 5 1

= - 2 - √5

By substituting;

2-√3 + √5+√3 -2-√5 = 0

(2,-2 cancels and √3,-√3 cancels and √5,-√5 cancels)

0 = 0