Huy Toàn 8A (TL)
28/07/2018 at 07:14-
Tôn Thất Khắc Trịnh 31/07/2018 at 08:25Simplifying? Okay
\(\dfrac{2}{x-1}+\dfrac{2x-1}{x^2+x+1}-\dfrac{x^2+6x+2}{1-x}=\dfrac{2}{x-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{x^2+6x+2}{x-1}\)Condition for x: \(x\ne1\)
\(\dfrac{2\left(x^2+x+1\right)}{x^3-1}+\dfrac{\left(2x-1\right)\left(x-1\right)}{x^3-1}+\dfrac{\left(x^2+6x+2\right)\left(x-1\right)}{x^3-1}\)
\(\Rightarrow2x^2+2x+2+2x^2-2x-x+1+x^3+6x^2+2x-x^2-6x-2\)
\(=x^3+3x^2-5x+1\)
(I can't factorize that polynomial
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Huy Toàn 8A (TL) 30/07/2018 at 02:10
Simplify the expression
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Huy Toàn 8A (TL) 29/07/2018 at 01:32
You should give the same denominator
Later Addition numerator
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Tôn Thất Khắc Trịnh 28/07/2018 at 17:34
Sorry but... I have no idea how to calculate that when you haven't given us any values of x to work with...