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simplify using only positive exponents. do not rationalize the denominator.

1. square root (4x-16) / fourth root {(x-4)^3}

2. [ (1 / x^-2) + (4 / x^-1 y^-1) + (1 / y^-2) ]^(-1/2)



sorry if its confusing (specially the second one...) let me know if you want clarification on any part of it!! thanks for the help!!!


please show steps...i think i have the right answer for 2, but im not sure... i got 1/x + 4/sqrtxy + 1/y

2007-09-02 08:35:22 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

1.) square root (4x-16) / fourth root {(x-4)³}

pull out a 4 first from the sqrt(4x-16)
sqrt(4)sqrt(x-4) so we have

[2 * sqrt(x-4)]/((x-4)^3)^1/4

((x-4)^3)^(1/4) = (x-4)^(3*(1/4)) = (x-4)^(3/4)

now our problem is : 2 * (x-4)^(1/2) / (x-4)^(3/4)
when dividing with exponents you subtract exponents so
our problem can simplify to:

2 * (x-4)^((1/2)-(3/4)) = 2 * (x-4)^(-1/4) = 2 / (x-4)^(1/4)

2 divided by the fourth root of the quantity 'x-4'

2.) [ (1 / x^-2) + (4 / x^-1 y^-1) + (1 / y^-2) ]^(-1/2)

All the components in this problem are raised to negative power which means you can each of those variables and move them to the opposite side of the divisor:

so 1/x^-2 = x²
4/x^-1y^-1 = 4xy
(1 / y^-2) = y²

as they are they cannot be combined since there is no common terms. to finish the whole thing is raised to the negative 1/2 power which means 1 divided by the square root of the thing.

1 / sqrt(x² + 4xy + y²)

this thing cannot be factored so easily holding y constant factors into (-4y ± sqrt(12y²))/2 so not much chance for making this any simpler.

2007-09-06 19:04:16 · answer #1 · answered by z32486 3 · 0 0

Let`s have a go!
Question 1
2 ( x - 4 ) ^ (1 / 2) / ( x - 4 ) ^ (3 / 4)
2 ( x - 4 ) ^ ( - 1 / 4)
2 / ( x - 4) ^ (1 / 4)

Question 2
( x ² + 4 x y + y ² ) ^ ( - 1 / 2)
1 / ( x ² + 4 x y + y ² ) ^ (1 / 2)

2007-09-06 11:00:13 · answer #2 · answered by Como 7 · 0 0

Put everything under the 4th root and factor the 4:

4th root( [16(x-4)^2] / (x-4)^3 )
(2) 4th root( 1 / (x-4) )

Clarify the second

2007-09-02 15:50:43 · answer #3 · answered by 037 G 6 · 0 0

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