这道题该怎么算啊？？求x
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原式={[(√x+1)^2-(√x-1)^2]/(x-1)+4√x}-[(x-1)/√x]
=[4√x//(x-1)+4√x]-[(x-1)/√x]
=4√x[1/(x-1)+1]-[(x-1)/√x]
=4x√x/(x-1)-[(x-1)/√x]
=[4x^2-(x-1)^2]/(x-1)√x
=(3x^2+2x-1)/(x-1)√x
=(3x+1)(x-1)/(x-1)√x
=(3x+1)/√x
=[(3x+1)√x]/x
=[4√x//(x-1)+4√x]-[(x-1)/√x]
=4√x[1/(x-1)+1]-[(x-1)/√x]
=4x√x/(x-1)-[(x-1)/√x]
=[4x^2-(x-1)^2]/(x-1)√x
=(3x^2+2x-1)/(x-1)√x
=(3x+1)(x-1)/(x-1)√x
=(3x+1)/√x
=[(3x+1)√x]/x
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