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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“分数化为小数”: \(\frac{3}{2} = \)___\(\frac{4}{{11}} = \)___.



知识点:第三章 一元一次方程


参考答案:1.5;\(0.\mathop 3\limits^. \mathop 6\limits^. \)

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