正常的分数乘法：

$\cfrac{3}{4} \times \cfrac{4}{3} = \cfrac{1 \atop \bcancel{3} }{\bcancel{4} \atop 1} \times \cfrac{1 \atop \bcancel{4} }{\bcancel{3} \atop 1} = 1$

我的分数乘法：

$\cfrac{3}{4} \times \cfrac{4}{3} = \cfrac{\bcancel{3}}{\bcancel{4}} \times \cfrac{\bcancel{4}}{\bcancel{3}} = 1$

正常的解方程：

$$\begin{aligned} 1+ \cfrac{3}{4} x &= 4\\ 1+ \cfrac{3}{4} x-1 &= 4-1\\ \cfrac{3}{4}x &= 3\\ \cfrac{3}{4}x \div \cfrac{3}{4} &= 3 \div \cfrac{3}{4}\\ x &= 3 \times \cfrac{4}{3}\\ x &= {1 \atop \bcancel{3}} \times \cfrac {4}{\bcancel{3}\atop 1}\\ x &= 4 \end{aligned}$$

我的解方程：

$$\begin{aligned} 1+ \cfrac{3}{4} x &= 4\\ \cfrac{3}{4} x &= 4-1\\ x &= 3 \times \cfrac{4}{3}\\ x &= 4 \end{aligned}$$

想问一下我的方法会不会在一些大考试上出事