Can't argue with success? Looking for "bad math" that "gets away with it"

I was quite amused when a student produced the following when cancelling a fraction:

$$\frac{x^2-y^2}{x-y}$$

He began by "cancelling" the $x$ and the $y$ on top and bottom, to get:

$$\frac{x-y}{-}$$

and then concluded that "two negatives make a positive", so the final answer has to be $x+y$.

Here's a pretty funny one from xkcd.

A student in a test was asked to give an example of two irrational numbers whose sum is irrational.

He chose $x = \sqrt{2}$, and $y=\sqrt{3}$, and computed the sum $x+y$ using a calculator. Unfortunately, he only took two digits, which led to the following:

$x = 1.41$, and $y = 1.73$, which implies that $x+y = 3.14$.

The student concluded that $\sqrt{2}+\sqrt{3}=\pi$.

Here's another classical freshman calculus example:

Find $\frac{d}{dx}x^x$.

Alice says "this is like $\frac{d}{dx}x^n = nx^{n-1}$, so the answer is $x x^{x-1} = x^x$." Bob says "no, this is like $\frac{d}{dx}a^x = \log a \cdot a^x$, so the answer is $\log x \cdot x^x$." Charlie says "if you're not sure, just add the two terms, so you'll get partial credit".

The answer $\frac{d}{dx}x^x = (1 + \log x)x^x $ turns out to be correct.