题目: 化简代数式(x2+4x-4)÷x2-4x2+2x,当x满足x-32(2x-1)≤4 ,①1+3x2>2x-1 ,②且为正整数时,求代数式的值. 答案: 原式=x2-4x+4x÷(x+2)(x-2)x(x+2)=(x-2)2x•xx-2=x-2.其中x≠0且x≠±2.解不等式组得:-54≤x<5,且x为正整数.∴x=1.∴原式=-1.
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