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Arithmetic · real student question

解方程求 X:(X − 40.354/14.4) × 3562/3600 = (147.304 × 600 + 40.354 × 1492)/(14.4 × 3600)。

Question

解方程求 XX:

(X−40.35414.4)⋅35623600=147.304⋅600+40.354⋅149214.4⋅3600.\left(X-\frac{40.354}{14.4}\right)\cdot\frac{3562}{3600}=\frac{147.304\cdot 600+40.354\cdot 1492}{14.4\cdot 3600}.

Step-by-step solution

  1. 先把左边那两个常数算出来。

    40.35414.4=2.8023611111…,35623600=0.9894444444…\frac{40.354}{14.4}=2.8023611111\ldots,\qquad \frac{3562}{3600}=0.9894444444\ldots

    两者都是精确的有理数,所以把它们保留成分数(4035414400\tfrac{40354}{14400} 和 17811800\tfrac{1781}{1800})而不是截断的小数,可以避免误差在后面那次除法里累积起来。

  2. 算右边的分子,要小心。

    147.304×600=88,382.4,40.354×1492=60,208.168.147.304\times 600=88{,}382.4,\qquad 40.354\times 1492=60{,}208.168.

    第二个乘积是需要复核的那一个:40.354×1492=40.354×1500−40.354×8=60,531−322.832=60,208.16840.354\times1492=40.354\times1500-40.354\times8=60{,}531-322.832=60{,}208.168。两者之和为

    88,382.4+60,208.168=148,590.568.88{,}382.4+60{,}208.168=148{,}590.568.

  3. 算分母,再算出右边的数值。

    14.4×3600=51,840,148,590.56851,840=2.8663304012…14.4\times 3600=51{,}840,\qquad \frac{148{,}590.568}{51{,}840}=2.8663304012\ldots

  4. 把那次乘法还原掉。 两边同除以 35623600\tfrac{3562}{3600}:

    X−2.8023611111=2.86633040120.9894444444=2.8969088839.X-2.8023611111=\frac{2.8663304012}{0.9894444444}=2.8969088839.

  5. 把那次减法也还原掉,并给出精确值。

    X=2.8969088839+2.8023611111=5.6992699950.X=2.8969088839+2.8023611111=5.6992699950.

    若整个计算全程用精确分数进行,得到 X=73,082,87912,823,200X=\dfrac{73{,}082{,}879}{12{,}823{,}200},其小数展开是 5.699269995009…5.699269995009\ldots —— 与走小数路线的结果在十位小数上一致。顺带一提,若把 40.354×149240.354\times1492 误算成 60,192.56860{,}192.568(一个常见的手误),右边就会变成 2.86602952.8660295,答案会变成 5.69896595.6989659,第四位小数就错了。

Answer

X=73,082,87912,823,200≈5.699270X=\frac{73{,}082{,}879}{12{,}823{,}200}\approx 5.699270

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