#### Answer

The value of the expression $\frac{x-1}{x+3}\le 0$ is $(-3,1]$

#### Work Step by Step

We know that for the inequality $\frac{x-1}{x+3}\le 0$, two conditions are possible:
$\left( x-1 \right)\le 0$ And $\left( x+3 \right)\gt0$
Or
$\left( x-1 \right)\ge 0$ And $\left( x+3 \right)\lt0$
There is no equal sign with the expression $ x+3$ since at $ x=-3$ the function does not exist.
Now, from the first condition, we have $ x\le 1$ and $ x>-3$.
From the second condition, we have $ x\ge 1$ and $ x-3$.
Thus, the solution for x is $(-3,1]$.