13.某班在体育课上进行1000米测试,在起点处学生小明比小华先跑1分钟,当小明到达终点时,小华还有440米没跑.已知小明每秒钟比小华每秒钟多跑1米.设小华速度为x米/秒,则可列方程为( )
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| A. | $\frac{1000}{x+1}$+1=$\frac{1000-440}{x}$ | B. | $\frac{1000}{x+1}$+60=$\frac{1000-440}{x}$ | ||
| C. | $\frac{1000}{x+1}$-1=$\frac{1000-440}{x}$ | D. | $\frac{1000}{x+1}$-60=$\frac{1000-440}{x}$ |