Solve the following equations:
a) $\frac{4 x-1}{6}-\frac{5 x-\frac{2}{3}}{10}+\frac{9-\frac{x}{2}}{3}=\frac{101}{20}$
b) $\frac{x}{2}-\frac{2 x-\frac{10-7 x}{3}}{2}+\frac{x-\frac{1+x}{3}}{3}=\frac{14}{9}$
c) $1+\frac{1+\frac{1+\frac{1+\frac{x}{5}}{4}}{3}}{2}=2-2 x-\frac{7-\frac{241 x}{5}}{24}$.
a) Multiplying both sides of the equality by the least common multiple of the denominators, 60:

$$
40 x-10-30 x+4+180-10 x=303
$$

from which, after rearrangement and addition,

$$
0=129
$$

This is an impossibility (or "contradiction"), which means there is no value of $x$ that satisfies our requirements. (Do not say it is an inequality, as that can be misleading).

b) Multiplying both sides of the equality by 18:

$$
9 x-18 x+30-21 x+6 x-2-2 x=28
$$

from which, after rearrangement and simplification,

$$
-26 x=0
$$

from which

$$
x=0
$$

(Thus, this is an equation whose only root is $x=0$. Do not say there is no solution, because 0 is just as much a root as, for example, $\frac{2}{3}$ or -5).

c) Simplifying the left side:

$$
1+\frac{1+\frac{1+\frac{5+x}{20}}{3}}{2}=1+\frac{1+\frac{20+5+x}{60}}{2}=1+\frac{60+25+x}{120}=\frac{205+x}{120}
$$

so our equality becomes

$$
\frac{205+x}{120}=2-2 x-\frac{35-241 x}{120}
$$

or

$$
205+x=240-240 x-35+241 x
$$

from which, after rearrangement and simplification,

$$
0=0
$$

This is an identity, meaning any value of $x$ satisfies our requirements. It is incorrect to say "the solution is 0". We also should not say "there is no solution", because the original equality holds for any value of $x$. We can say: "there is no definite solution".