x/(4x²-4x+1)÷1/(2x²-x)=?x 1_________ ÷ _________ = (4x²-4x+1) (2x²-x)
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![x/(4x²-4x+1)÷1/(2x²-x)=?x 1_________ ÷ _________ = (4x²-4x+1) (2x²-x)](/uploads/image/z/1344893-5-3.jpg?t=x%2F%EF%BC%884x%26%23178%3B-4x%2B1%EF%BC%89%C3%B71%2F%EF%BC%882x%26%23178%3B-x%EF%BC%89%3D%3Fx+1_________+%C3%B7+_________+%3D+%EF%BC%884x%26%23178%3B-4x%2B1%EF%BC%89+%EF%BC%882x%26%23178%3B-x%EF%BC%89)
x/(4x²-4x+1)÷1/(2x²-x)=?x 1_________ ÷ _________ = (4x²-4x+1) (2x²-x)
x/(4x²-4x+1)÷1/(2x²-x)=?
x 1
_________ ÷ _________ =
(4x²-4x+1) (2x²-x)
x/(4x²-4x+1)÷1/(2x²-x)=?x 1_________ ÷ _________ = (4x²-4x+1) (2x²-x)
x/(4x²-4x+1)=x/(2x-1)²,1/(2x²-x)=1/[x(2x-1)],因此原式=[x/(2x-1)²]÷{1/[x(2x-1)]=[x*x(2x-1)]/(2x-1)²=x²/(2x-1)
x²/2x-1