Math

I was wondering if I solved these problems right?
Simplify the following:

3xy^2 4xz
______ + ______ = 7xyz
2y 3x _____
5


2(x-1) 3(x^2-4)
______ * ________ = 6(x-2)
4(x+2) 5x-5 ______
20

x^2-x-6 3(x^2-4)
________ / __________ = (x-3) (x+5)
7x+7 x^2+6x+5 ______________
21(x-2)

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  1. Wow, did you look how your typing showed up ??

    I will take a "guess" what the last one is ....
    (x^2 - x - 6)/(7x+7) ÷ 3(x^2-4)/(x^2+6x+5) = (x-3)(x+5)/(21(x-2))

    (x-3)(x+2)/(7(x+1)) (x+1)(x+5)/(3(x+2)(x-2)) = (x-3)(x+5)/(21(x-2))

    (x-3)(x+5)/(21(x-2) = (x-3)(x+5)/(21(x-2))
    everything cancels for
    1 = 1

    so the equation is an identity and true for all values of x, except x ≠ -1, ± 2,

    retype the others using brackets.

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  2. 1)

    (3xy^2)/(2y) + (4xz)/(3x)

    2)

    (2(x-1))/(4(x+2)) *(3(x^2-4))/(5x-5)

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