一道初三的化简求值题目，求大神解答。重赏~~~

原式=[(3x+4)/(x²-1)-2(x+1)/(x²-1)]+(x+2)/(x-1)²
=(3x+4-2x-2)/(x²-1)+(x+2)/(x-1)²
=(x+2)/(x²-1)+(x+2)/(x-1)²
=(x+2)(x-1)/(x+1)(x-1)²+(x+2)(x+1)/(x+1)(x-1)²
=(x+2)(x-1+x+1)/(x+1)(x-1)²
=2x(x+2)/(x+1)(x-1)²

x+4>0，得：x>-4
2x+5<1，得：x<-22
所以，-4整岩早数解为饥搭：x=-3
把x=-3代入2x(x+2)/(x+1)(x-1)²
得：原式=2x(x+2)/(x+1)(x-1)²=-3/16

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不等式组x+4＞0；2x+5＜1的整数解派扰悔。-4，[(3x+4）/（x²-1）-2/（x-1）]+[（x+2）/（x²-2x+1）]
=[(3*-3+4）/（-3)²-1）-2/（-3-1）]+[（-3+2）/（-3)²-2*(-3)+1）]
=[-5/尘正8+1/2]+[-1/(16)]
=-1/8-1/李纳16
=-3/16

不等式组的解是-4＜明顷x＜-2，x取整数，所以x=-3。
带入式胡链子化简激做陆：-5/8+1/2-1/16=-3/16