当考试最后一题是:如何证明1+1=3?
  • 板块灌水区
  • 楼主tujize
  • 当前回复2
  • 已保存回复2
  • 发布时间2024/12/28 14:14
  • 上次更新2024/12/28 14:25:47
查看原帖
当考试最后一题是:如何证明1+1=3?
1255429
tujize楼主2024/12/28 14:14

2=422=4-2

2=4022=4-0-2

2=492+9222=4-\displaystyle\frac{9}{2}+\displaystyle\frac{9}{2}-2

2=(492)2+9222=\sqrt{(4-\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=422×4×92+(92)2+9222=\sqrt{4^2-2\times4\times \displaystyle\frac{9}{2}+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=424×9+(92)2+9222=\sqrt{4^2-4\times9+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=1636+(92)2+9222=\sqrt{16-36+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=20+(92)2+9222=\sqrt{-20+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=2545+(92)2+9222=\sqrt{25-45+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=522×5×92+(92)2+9222=\sqrt{5^2-2\times5\times\displaystyle\frac{9}{2}+(\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=(592)2+9222=\sqrt{(5-\displaystyle\frac{9}{2})^2}+\displaystyle\frac{9}{2}-2

2=592+9222=5-\displaystyle\frac{9}{2}+\displaystyle\frac{9}{2}-2

2=522=5-2

2=32=3

1+1=31+1=3


🙂

2024/12/28 14:14
加载中...