当考试最后一题是:如何证明1+1=3?
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  • 楼主tujize
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  • 发布时间2024/12/28 14:14
  • 上次更新2024/12/28 14:25:47
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当考试最后一题是:如何证明1+1=3?
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tujize楼主2024/12/28 14:14

2=4−22=4-2

2=4−0−22=4-0-2

2=4−92+92−22=4-\displaystyle\frac{9}{2}+\displaystyle\frac{9}{2}-2

2=(4−92)2+92−22=\sqrt{(4-\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=42−2×4×92+(92)2+92−22=\sqrt{4^2-2\times4\times \displaystyle\frac{9}{2}+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=42−4×9+(92)2+92−22=\sqrt{4^2-4\times9+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=16−36+(92)2+92−22=\sqrt{16-36+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=−20+(92)2+92−22=\sqrt{-20+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=25−45+(92)2+92−22=\sqrt{25-45+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=52−2×5×92+(92)2+92−22=\sqrt{5^2-2\times5\times\displaystyle\frac{9}{2}+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=(5−92)2+92−22=\sqrt{(5-\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=5−92+92−22=5-\displaystyle\frac{9}{2}+\displaystyle\frac{9}{2}-2

2=5−22=5-2

2=32=3

1+1=31+1=3


🙂

2024/12/28 14:14
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