Question

先化简，再求值：[x+2/（x^2-2x）-(x-1）/（x^2-4x+4)]÷2x-8/(x^3-2x^2)

在满足不等式组-x/2≤0 1/3(x-2)≤2的整数解中选择你喜欢的数求出代数式的值

Answer 1

解：原式=[(x+2)/(x^2-2x) - (x-1)/(x^2-4x+4)]÷2(x-4)/(x^3-2x^2)
=[(x+2)/(x^2-2x) - (x-1)/(x-2)²)]* [x²(x-2)]/2(x-4)
=[(x+2)- x(x-1)/(x-2)]*x/2(x-4)
=[(x²-4-x²+x)/(x-2)]*x/2(x-4)
=[(x-4)/(x-2)]*x/2(x-4)
=x/2(x-2)

x≥0
x-2≤6
x≤8
解集为0≤x≤8
x=1时，原式=1/2*(-1)=-1/2