# Thread: Review Problems (Test Tommorow)

1. ## Review Problems (Test Tommorow)

I've been reviewing all day for a test tommorow and the following are the problems I have no idea how to complete. I don't expect them all to be answered but any help/explanation would be greatly appreciated.

1) Simplify the expression.
$\displaystyle \frac{3(1+x)^1/^3-x(1+x)^-2/^3}{(1+x)^2/^3}$

^Sorry about the messed up exponents.

2) Find all real solutions to the equation.
$\displaystyle \frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$

Thanks.

2. Originally Posted by Ballplaya4237
I've been reviewing all day for a test tommorow and the following are the problems I have no idea how to complete. I don't expect them all to be answered but any help/explanation would be greatly appreciated.

1) Simplify the expression.
$\displaystyle \frac{3(1+x)^1/^3-x(1+x)^-2/^3}{(1+x)^2/^3}$

^Sorry about the messed up exponents.

2) Find all real solutions to the equation.
$\displaystyle \frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$

Thanks.
for the first one, i couldnt really understand, but i can give you tip to replace (1+x) with a variable, and it'll look a lot cleaner, then sub it back in at the end.

im kinda busy so for the second ill just tell you that u need restrictions from the denominators at the end, for ex. x=/=-5 because it would make that first fraction divided by zero.

3. Originally Posted by Ballplaya4237
2) Find all real solutions to the equation.
$\displaystyle \frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$

Thanks.
$\displaystyle (x^2 - 4) = (x - 2)(x + 2)$

Multiply straight through with $\displaystyle (x - 2)(x + 2)$

Then you get:

$\displaystyle (x+5)(x+2) = (5)(x-2)+28$

$\displaystyle x^2 + 7x + 10 = 5x - 10 + 28$

$\displaystyle x^2 + 7x - 5x -28 = 0$

$\displaystyle x^2 + 2x -28 = 0$

$\displaystyle x = 4,3851...$

OR

$\displaystyle x = -6,3851...$

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Still working on the first one.

4. Originally Posted by Ballplaya4237
1) Simplify the expression.
$\displaystyle \frac{3(1+x)^1/^3-x(1+x)^-2/^3}{(1+x)^2/^3}$
Let $\displaystyle (1 + x)^{ \frac{1}{3} } = y$

$\displaystyle = \frac{ 3y - xy^{-2} }{ y^2 }$

$\displaystyle = \frac{ y(3 - xy^{-3}) }{ y(y) }$

$\displaystyle = \frac{ (3 - xy^{-3}) }{ (y) }$

Now set $\displaystyle y = (1 + x)^{ \frac{1}{3} }$ back into the equation.

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Good luck with the test.