(-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)]
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(-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)](-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)](-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)](-x^
(-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)]
(-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)]
(-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)]
(-x^2n-2)·(-x)^5÷[x^n+1·x^n·(-x)]
=(x^2n-2)·(x)^5÷[-x^n+1·x^n+1]
=-x^(2n-2+5)÷x^(2n+2)
=-x^(2n+3-2n-2)
=-x
原式是这样么?
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(-x^2n-2)(-x)^5÷[x^n+1·x^n`(-x)]
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