分式求值是分式运算中的一类常见问题,对计算能力的要求较高。在求解此类问题时,既要注意基本法则的应用,也要掌握相关的解题技巧。下面举例说明。一、整体通分3例1计算x2+x+1-x3/x-1分析:把(x2+x+1)看成一个整体,对其进行通分,并且分子还可利用乘法公式简化运算。解:原式=(x-1)(x2+x+1)-x3=x3-1-x3=-x-1x-1x-11。x-1二、部分通分例2计算:1-1-2-4x-1x+1x2+1x4。+1分析:按照常规解法是把四个分母一起通分,这样求解过于繁琐。若选择前面两个分式通分,然后再逐个通分,这样化繁琐为简单。解%原式=2-2-4(x+1)(x-1)x2+1x4=+1 Fractional evaluation is a common type of problem in fractional computing, requiring a high level of computational power. In solving these problems, we should not only pay attention to the application of the basic law, but also master the relevant problem-solving skills. Here’s an example. First, the overall pass in three cases 1 Calculate x2 + x + 1-x3 / x-1 Analysis: The (x2 + x +1) as a whole, pass through them, and the numerator can also use the multiplication formula to simplify the operation . Solution: Original = (x-1) (x2 + x + 1) -x3 = x3-1-x3 = -x-1x-1x-11. x-1 Second, the partial pass Example 2 Calculation: 1-1-2-4x-1x + 1x2 + 1x4. +1 Analysis: According to the conventional solution is to pass through the four denominators, this solution is too complicated. If you choose the first two fractional points, and then one by one, so cumbersome as simple. Solution% Original = 2-2-4 (x + 1) (x-1) x2 + 1x4 = + 1