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初中数学七年级上册(581题)


阅读理解:仔细阅读下列材料:

我们学习实数后知道:“分数均可化为有限小数或无限循环小数”. 反之,“有限小数或无限循环小数均可化为分数”. 

例如:\(\frac{1}{4} = 1 \div 4 = 0.25\)\(1\frac{3}{5} = 1 + \frac{3}{5} = 1 + 0.6 = 1.6\)\(1\frac{3}{5} = \frac{8}{5} = 8 \div 5 = 1.6\)\(\frac{1}{3} = 1 \div 3 = 0.\dot 3\)

反之,\(0.25 = \frac{{25}}{{100}} = \frac{1}{4}\)\(1.6 = 1 + 0.6 = 1 + \frac{6}{{10}} = 1\frac{3}{5}\)\(1.6 = \frac{{16}}{{10}} = \frac{8}{5}\)

那么\(0.\dot 3\)怎么化为\(\frac{1}{3}\)呢?

解:\(\because 0.\dot 3 \times 10 = 3.\dot 3 = 3 + 0.\dot 3\)

\(\therefore \)不妨设\(0.\dot 3 = x\),则上式变为\(10x = 3 + x\),解得\(x = \frac{1}{3}\) \(0.\dot 3 = \frac{1}{3}\)

根据以上材料,回答下列问题.

将小数\(1.\mathop {15}\limits^{\bullet \bullet } \)化为分数,请写出推理过程.



知识点:第三章 一元一次方程


参考答案:\(1.\mathop 1\limits^. \mathop 5\limits^. = 1 + 0.\mathop 1\limits^. \mathop 5\limits^. \),∵\(0.\mathop 1\limits^. \mathop 5\limits^. \times 100 = 15.\mathop 1\limits^. \mathop 5\limits^. = 15 + 0.\mathop 1\limits^. \mathop 5\limits^. \)

设\(0.\mathop 1\limits^. \mathop 5\limits^. = x\),则上式变为\(100x = 15 + x\)

解得\(x = \frac{{15}}{{99}} = \frac{5}{{33}}\),所以\(1.\mathop 1\limits^. \mathop 5\limits^. = 1 + 0.\mathop 1\limits^. \mathop 5\limits^. \)\( = 1 + \frac{5}{{33}} = \frac{{38}}{{33}}\)

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