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domain of sqrt{x^2-3x}/(x^2-6x+5)-x/(x^3-8)
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hmmmm. solve for x^2 -3x = 0. im... thinking... x is 3 or above.
you mean \[x^2-3x \ge0\] and then x(x-3) and then draw a table?
x2 -6x+5=(x-1)(x-5) x not can being equal 1 and 5 x3 -8 --- 2 on exponent3 -8=0 so resulted that x not can being 2 so the domain for this equation will be from -infinity till 1 without 1 and from 1 till 5 without 1,2 and 5 and from 5 till + infinity without 5
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