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九年级数学solution一般
题目
先化简,再求值:(x+3x2−x−xx2−2x+1)÷2x−3x(\frac{x+3}{x^2-x}-\frac{x}{x^2-2x+1})÷\frac{2x-3}{x},其中xx满足x2−2x−4=0x^{2}-2x-4=0.
知识点：分式的化简求值章节：未标注

答案与解析

答案

(x+3x2−x−xx2−2x+1)÷2x−3x(\frac{x+3}{x^2-x}-\frac{x}{x^2-2x+1})÷\frac{2x-3}{x}
=[(x+3)(x−1)x(x−1)2−x2(x−1)2]⋅x2x−3=[\frac{(x+3)(x-1)}{x(x-1)^{2}}-\frac{x^{2}}{(x-1)^{2}}]\cdot \frac{x}{2x-3}
=x2+2x−3−x2x(x−1)2⋅x2x−3=\frac{x^{2}+2x-3-x^{2}}{x(x-1)^{2}}\cdot \frac{x}{2x-3}
=1x2−2x+1=\frac{1}{x^{2}-2x+1}.
∵x2−2x−4=0\because x^{2}-2x-4=0，
∴x2−2x=4\therefore x^{2}-2x=4，
∴\therefore原式=14+1=15=\frac{1}{4+1}=\frac{1}{5}.

解析

(x+3x2−x−xx2−2x+1)÷2x−3x(\frac{x+3}{x^2-x}-\frac{x}{x^2-2x+1})÷\frac{2x-3}{x}
=[(x+3)(x−1)x(x−1)2−x2(x−1)2]⋅x2x−3=[\frac{(x+3)(x-1)}{x(x-1)^{2}}-\frac{x^{2}}{(x-1)^{2}}]\cdot \frac{x}{2x-3}
=x2+2x−3−x2x(x−1)2⋅x2x−3=\frac{x^{2}+2x-3-x^{2}}{x(x-1)^{2}}\cdot \frac{x}{2x-3}
=1x2−2x+1=\frac{1}{x^{2}-2x+1}.
∵x2−2x−4=0\because x^{2}-2x-4=0，
∴x2−2x=4\therefore x^{2}-2x=4，
∴\therefore原式=14+1=15=\frac{1}{4+1}=\frac{1}{5}.

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