Filling in a missing portion of a truth table I have the following truth table:
$$
\boxed{
\begin{array}{c|c|c|c}
a & b & c & x \\ \hline
F & F & F         & F \\
F & F & T         & F \\
F & T & F         & F \\
F & T & T         & F \\
T & F & F         & T \\
T & F & T         & T \\
T & T & F         & F \\
T & T & T         & T 
\end{array}}
$$
I am supposed to use any logical operator: and, or, not. I tried like $10$ formulas already, and I am not able to think of what combination of $a,b,c$ will get that $x$ column right. Have I missed something really silly? 
 A: $x$ is only True in rows 5, 6, and 8, all of which have $a$ as True.  This means that we should look for:
$x$ = $a$ ∧ something
Whatever that "something" is, it should return True for all combinations of $b$ and $c$ except for $b$ True and $c$ False (which was row 7, where $x$ is False).
A: because there are fewer $f$ than $t$'s, i will look at 
$$\begin{align}x &= ab'c'+ab'c++abc\\
&=ab'(c+c')+abc\\
&=ab'+abc\\
&=a(b'+bc)\\
&=a(b'+c)\end{align}$$
A: You need to think about what it means to find a formula $x$ that fits the truth table. The easiest way to generate such a formula is simply to write a disjunction of all the cases where you want $x$ to be true.
Essentially, $x$ will say "I am true when this combination of $A$, $B$ and $C$ occurs, or when that one occurs, or when that one occurs, etc.". What you have to do is look up the lines where $x$ is true, get the corresponding truth values of $A$, $B$ and $C$, connect them with a conjunction, and link everything up with disjunctions.
By looking at the three lines where $x$ is true, you get:
$$x:=(A\land\neg B\land\neg C)\lor(A\land\neg B\land C)\lor(A\land B\land C)$$
Make sure you understand what this means and how the truth table follows.
Now, any other formula fitting the truth table will be equivalent to $x$. For instance, we can "improve" $x$ by noticing that it requires $A$ to always be true, such that we can take it out of the parentheses:
$$x=A\land\big((\neg B\land\neg C)\lor(\neg B\land C)\lor(B\land C)\big)$$
The part inside the big brackets is listing three out of four possible cases for the truth values of the pair $(B,C)$. In fact, the missing case is $(B\land\neg C)$, so that we can rewrite this segment to say "any combination of $B$ and $C$ works, except for $(B\land\neg C)$. We do so using a negation:
$$x=A\land\neg(B\land\neg C)$$
And this is the most succinct expression of $x$ using only the connectives $\left\{\land,\neg\right\}$. Furthermore, you can compare it to the truth table: $x$ is true whenever $A$ is true except for the specific case where $B$ is true and $C$ is false. This corresponds precisely to the last four lines of the table, without the penultimate.
A: Hint: 

 $$x\equiv a\land(\neg b\lor c).$$

How did I come up with this? Well, if $a$ is false, then any conjunction with $a\land(\text{something here})$ will be false as well. That takes care of the top half of the truth table. All that is left, then, is to figure out how to manipulate a combination of $b$ and $c$ to combine with $a$ to give the desired truth values. Thus, what you want is what is given above (grayed out in case you want to try to deduce it yourself).
