初中数学七年级易错母题017：有理数加减拓展提升练习例题解析
A . 3 B . -3 C . 3或-9 D . -3或9
答案：D
解析：因为|(-3)+♦|=6,所以(-3)+♦=±6,故分情况讨论: ①当(-3)+♦=6时,得♦ =9; ②当(-3)+♦=-6时,得♦=-3.故♦的值可能是-3 或 9。
答案：\(\displaystyle \frac{1}{{12}}\)
解析：\(\displaystyle 3*\left( {-4} \right)=\frac{1}{3}-\frac{1}{4}=\frac{1}{{12}}\)
计算：\(\displaystyle -5\frac{5}{6}+\left( {-9\frac{2}{3}} \right)+17\frac{3}{4}+\left( {-3\frac{1}{2}} \right)\)
原式\(\displaystyle =\left[ {-5+\left( {-\frac{5}{6}} \right)} \right]+\left[ {\left( {-9} \right)+\left( {-\frac{2}{3}} \right)} \right]+\left( {17+\frac{3}{4}} \right)+\left[ {\left( {-3} \right)+\left( {-\frac{1}{2}} \right)} \right]\)
\(\displaystyle =\left( {-5} \right)+\left( {-9} \right)+17+\left( {-3} \right)+\left[ {\left( {-\frac{5}{6}} \right)+\left( {-\frac{2}{3}} \right)+\frac{3}{4}+\left( {-\frac{1}{2}} \right)} \right]\)
\(\displaystyle =0+-1\frac{1}{4}\)
\(\displaystyle =-1\frac{1}{4}\)
上面这种解题方法叫做拆项法
计算：\(\displaystyle \left( {-2023\frac{5}{6}} \right)+\left( {-2022\frac{2}{3}} \right)+4046\frac{2}{3}+\left( {-1\frac{1}{2}} \right)\)
答案：\(\displaystyle -1\frac{1}{3}\)
解析：
原式\(\displaystyle =\left[ {\left( {-2023} \right)+\left( {-\frac{5}{6}} \right)} \right]+\left[ {\left( {-2022} \right)+\left( {-\frac{2}{3}} \right)} \right]+\left[ {4046+\frac{2}{3}} \right]+\left[ {\left( {-1} \right)+\left( {-\frac{1}{2}} \right)} \right]\)
\(\displaystyle =\left[ {\left( {-2023} \right)+\left( {-2022} \right)+4046+\left( {-1} \right)} \right]+\left[ {\left( {-\frac{5}{6}} \right)+\left( {-\frac{2}{3}} \right)+\frac{2}{3}+\left( {-\frac{1}{2}} \right)} \right]\)
\(\displaystyle =0+\left( {-1\frac{1}{3}} \right)\)
\(\displaystyle =-1\frac{1}{3}\)
