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Find $ x \in \mathbb{Q} $ such that $ 1 + \frac{2}{[x]} + \frac{3}{{x - [x] }} = 4 {x}$. Initially I was thinking that is easy. If we separate the therms with integer part and fractionary part and consider the functions with fractionary part she is continous functions and we have a solution all the time but we don't know if it is rational. I was trying to consider and an equation of degree two but doesn't work.

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If we write $k = [x]$, then the equation translates to $1 + \frac 2 k + \frac 3 {x - k} = 4x$, which after simplification gives $$4k x^2 - (4k^2 + k + 2)x + k^2 - k = 0.$$ Since this quadratic equation in $x$ has a rational solution, its discriminant $\Delta$ must be the square of some rational number $r$ (note that $k \neq 0$).

Thus we write $$r^2 = \Delta = (4k^2 + k + 2)^2 - 4 \cdot 4k \cdot (k^2 - k) = 16k^4 - 8k^3 + 33k^2 + 4k + 4.$$ Since $\Delta$ is an integer, we know that $r$ is also an integer, which we shall assumed to be non-negative.

Completing the square, we get $r^2 = (4k^2 - k + 4)^2 + 12k - 12$.

There are three cases: $k = 1$, $k > 1$, $k < 1$.

For $k = 1$, we solve the equation and get $x = 0$ or $\frac 7 4$. Obviously $x = \frac 7 4$ is a possible solution to the original equation.

If $k > 1$, then we have $r^2 > (4k^2 - k + 4)^2$ which gives $r \geq 4k^2 - k + 5$ (note that $4k^2 - k + 4$ is strictly positive for any $k$). Thus we have $$(4k^2 - k + 5)^2 \leq r^2 = (4k^2 - k + 4)^2 + 12k - 12$$ which leads to $8k^2 - 14k + 21 \leq 0$, which is impossible.

If $k < 1$, then a similar argument gives $r \leq 4k^2 - k + 3$ and hence $$(4k^2 - k + 3)^2 \geq r^2 = (4k^2 - k + 4)^2 + 12k - 12$$ which leads to $8k^2 + 10k - 5 \leq 0$. It is easy to see that the only integers $k$ satisfying this inequality are $k = 0, -1$, and solving the original equation confirms that no rational solution $x$ exists in this case.

Therefore the only rational solution is $x = \frac 7 4$.

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