3(1+x)sqrt 1/3 -x(1-x)sqrt -2/3
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(1+x) sqrt 2/3
please help
sqrt=square root?
yes
sqrt-2/3 ?
sorry i said it wrong. i just posted it again with correct info.
To simplify the given expression, we can follow a few steps:
Step 1: Distribute the numerator and the denominator.
3(1+x)sqrt(1/3) - x(1-x)sqrt(-2/3) is already distributed.
Step 2: Simplify the square root expressions separately in the numerator and denominator.
For the numerator:
sqrt(1/3) = sqrt(1)/sqrt(3) = 1/sqrt(3).
(Note: sqrt(1) is just 1)
For the denominator:
sqrt(-2/3) = sqrt((-2)/3) = sqrt(-2)/sqrt(3).
Step 3: Simplify the square roots.
For the numerator:
1/sqrt(3) cannot be simplified any further because sqrt(3) is the simplified form.
For the denominator:
We can simplify sqrt(-2) by writing it as i*sqrt(2) where i is the imaginary unit.
So, sqrt(-2)/sqrt(3) becomes (i*sqrt(2))/sqrt(3).
Step 4: Combine like terms.
Now that we have simplified the numerator and denominator separately, we can combine them:
[(3(1+x))/(1+x)] * [(1/sqrt(3))/((i*sqrt(2))/sqrt(3))]
Simplifying further:
3 * (1/sqrt(3))/(i*sqrt(2))/sqrt(3)
= 3 * (1/sqrt(3)) * (sqrt(3)/(i*sqrt(2)))
= (3 * 1 * sqrt(3))/(i * sqrt(2) * sqrt(3))
= 3 * sqrt(3) / i * sqrt(2 * 3)
= 3 * sqrt(3) / (i * sqrt(6))
Step 5: Rationalizing the denominator.
Since we have an imaginary number in the denominator, we need to rationalize it by multiplying both the numerator and denominator by the conjugate of the denominator, which is -i * sqrt(6).
Multiplying the numerator and denominator by -i * sqrt(6):
(3 * sqrt(3) / (i * sqrt(6))) * (-i * sqrt(6)/-i * sqrt(6))
= (-3i * sqrt(3) * sqrt(6)) / (i * i * sqrt(6) * sqrt(6))
= (-3i * sqrt(18)) / (-1 * 6)
= (-3i * sqrt(18)) / (-6)
= (3i * sqrt(18)) / 6
= (3i * sqrt(9 * 2)) / 6
= (3i * 3 * sqrt(2)) / 6
= (9i * sqrt(2)) / 6
= (3i * sqrt(2)) / 2
Therefore, the simplified expression is (3i * sqrt(2))/2.