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Math Help - Some expressions

  1. #1
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    Some expressions

    1. Simplify the expression
    (2a^-2-2b^-2)/(4a^-1+4b^-1)
    I know it goes to this
    (2/a^2-2/b^2)/(4/a+4/b)
    I eventually get it down to b-a/2ab
    But I guess that's incorrect? I think it has to do with how I'm making the denominators equal.

    2. add/subtract\
    (4x/x-1)+(4/1-x^2)-(2/x+1)
    don't know how to find the lcd?
    is the lcd (1+x)(1-x)(x-1)?

    thanks for help in advance.
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  2. #2
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    Hello, bilbobaggins!

    2. Add/subtract: . \frac{4x}{x-1} + \frac{4}{1-x^2} - \frac{2}{x+1}
    Factor a "minus" out of the middle denominator . . .

    We have: . \frac{4x}{x-1} - \frac{4}{x^2-1} - \frac{2}{x+1}

    Factor: . \frac{4x}{x-1} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\quad\hdots\;\;\text{LCD} \:=\: (x-1)(x+1)

    Convert: . \frac{4x}{x-1}\cdot{\color{blue}\frac{x+1}{x+1}} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\cdot{\color{blue}\frac{x-1}{x-1}}

    . . = \;\frac{4x(x+1) - 4 - 2(x-1)}{(x-1)(x+1)} \;=\;\frac{4x^2+4x-4-2x+2}{(x-1)(x+1)} \;=\;\frac{4x^2+2x-2}{(x-1)(x+1)}

    Factor: . \frac{2(2x^2+x-2)}{(x-1)(x+1)} \;=\;\frac{2(2x-1)(x+1)}{(x-1)(x+1)} \;=\; \frac{2(2x-1)}{x-1}

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  3. #3
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    Quote Originally Posted by bilbobaggins View Post
    1. Simplify the expression
    (2a^-2-2b^-2)/(4a^-1+4b^-1)
    I know it goes to this
    (2/a^2-2/b^2)/(4/a+4/b)
    I eventually get it down to b-a/2ab
    ...
    to #1.

    \left(2a^{-2}-2b^{-2}\right)\left(4a^{-1}+4b^{-1}\right)=8a^{-3}+8a^{-2}b^{-1} - 8a^{-1}b^{-2}-8b^{-3}

    The common denominator is a^3b^3. You'll get:

    \dfrac{8\left(-a^3+ab^2 -a^2b+b^3 \right)}{a^3b^3}=\dfrac{8\left(a(b^2-a^2)+b(b^2-a^2)  \right)}{a^3b^3}=\dfrac{8(a-b)(a+b)^2}{a^3b^3}

    which you can't simplify any more.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, bilbobaggins!

    Factor a "minus" out of the middle denominator . . .

    We have: . \frac{4x}{x-1} - \frac{4}{x^2-1} - \frac{2}{x+1}

    Factor: . \frac{4x}{x-1} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\quad\hdots\;\;\text{LCD} \:=\: (x-1)(x+1)

    Convert: . \frac{4x}{x-1}\cdot{\color{blue}\frac{x+1}{x+1}} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\cdot{\color{blue}\frac{x-1}{x-1}}

    . . = \;\frac{4x(x+1) - 4 - 2(x-1)}{(x-1)(x+1)} \;=\;\frac{4x^2+4x-4-2x+2}{(x-1)(x+1)} \;=\;\frac{4x^2+2x-2}{(x-1)(x+1)}

    Factor: . \frac{2(2x^2+x-2)}{(x-1)(x+1)} \;=\;\frac{2(2x-1)(x+1)}{(x-1)(x+1)} \;=\; \frac{2(2x-1)}{x-1}

    I'm guessing the 2 right here 2(2x^2+x-2) was a typo?

    Quote Originally Posted by earboth View Post
    to #1.

    \left(2a^{-2}-2b^{-2}\right)\left(4a^{-1}+4b^{-1}\right)=8a^{-3}+8a^{-2}b^{-1} - 8a^{-1}b^{-2}-8b^{-3}

    The common denominator is a^3b^3. You'll get:

    \dfrac{8\left(-a^3+ab^2 -a^2b+b^3 \right)}{a^3b^3}=\dfrac{8\left(a(b^2-a^2)+b(b^2-a^2)  \right)}{a^3b^3}=\dfrac{8(a-b)(a+b)^2}{a^3b^3}

    which you can't simplify any more.
    Wait, what? Did you distribute the numbers? Like multiplication? I know with division of fractions you're supposed to flip then multiply. It doesn't appear you did that. Could you further explain?

    Thanks for the help both of you.
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  5. #5
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    Quote Originally Posted by bilbobaggins View Post
    ...

    Wait, what? Did you distribute the numbers? Like multiplication? I know with division of fractions you're supposed to flip then multiply. It doesn't appear you did that. Could you further explain?

    Thanks for the help both of you.
    1. You have a product of sums therefore you have to use indeed the law of distribution.

    2. The summands contains powers therefore you have to use the rules of calculating powers.

    3. At this point I used the rules of adding fractions. I didn't write the fractions because I was pretty sure that you see that 8a^{-2}b^{-1}=8\cdot \dfrac1{a^2} \cdot \dfrac1{b}

    4. The followings steps are nothing but factor out common factors from partial sums.
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