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packjm_vuive

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nguyenbahiep1

[TEX]TXD: x \not= 1, x\not=-1 , x \not= 0 \\ \frac{(x+1)^2 -(x-1)^2}{(x-1)(x+1)} : \frac{x-1 +x+1}{x-1} = \frac{x-1}{2.(x+1)} \\ \frac{4x}{(x-1)(x+1)}.\frac{x-1}{2x} = \frac{x-1}{2.(x+1)}\\ \frac{2}{x+1} = \frac{x-1}{2.(x+1)} \\ \frac{4 -x +1}{2.(x+1)} = 0 \\ x = 5[/TEX]
 
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