[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)] 这个等于-1/2是怎么算的?

[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)] 这个等于-1/2是怎么算的?
[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)
[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)] 这个等于-1/2是怎么算的?

[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)[(x²-1)²+(2x+1)²-(x²+x+1)²]/[2(x²-1)(2x+1)] 这个等于-1/2是怎么算的?
其实你看到这个题目和答案,就可以想到,通过-(x²+x+1)²来合并拆分,与分母相互约去.
或者更简单,不用想的就是将分子每个都拆开,然后与分母相约.