How do you simplify #1/{1+sqrt(3)-sqrt(5)}#?
You're going to have to do a little work here to simplify this expression.
The way to go is by rationalizing the denominator. The only problem is the fact that your denominator is a trinomial, and conjugates are only formed for binomials.
More specifically, you get the conjugate of a binomial by changing the sign of its second term.
This means that you're going to have to group the denominator as a binomial, for which you can write
The denominator can be rewritten as
This, in turn, will be equal to
The expression becomes
Now do the same thing with the new denominator, i.e. find its conjugate
The denominator will be equal to
The numerator will be
The simplified expression will thus be
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To simplify the expression 1/{1+sqrt(3)-sqrt(5)}, we can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is 1+sqrt(3)+sqrt(5).
By doing this, we get: 1/{1+sqrt(3)-sqrt(5)} * (1+sqrt(3)+sqrt(5))/(1+sqrt(3)+sqrt(5))
Simplifying the numerator, we have: 1 + sqrt(3) + sqrt(5)
Expanding the denominator, we have: (1+sqrt(3)-sqrt(5))(1+sqrt(3)+sqrt(5)) = 1 + sqrt(3) + sqrt(5) + sqrt(3) + 3 - sqrt(15) - sqrt(5) - sqrt(15) + 5 = 10 - 2sqrt(15)
Therefore, the simplified expression is: 1/{1+sqrt(3)-sqrt(5)} = (1 + sqrt(3) + sqrt(5))/(10 - 2sqrt(15))
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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