Find the value of $x$

$\dfrac{1}{x-3}+\dfrac{1}{x-4}=\dfrac{1}{x-2}+\dfrac{1}{x-5}\\ \dfrac{x-4+x-3}{(x-3)(x-4)} = \dfrac{x-5+x-2}{(x-2)(x-5)}\\ (2x-7)(x-2)(x-5) = (2x-7)(x-3)(x-4)\\ (2x-7)(x^2-7x+10) - (2x-7)(x^2-7x+12) = 0\\ (2x-7)\left((x^2-7x+10) - (x^2-7x+12)\right) = 0\\ (2x-7)(-2) = 0\\ 2x-7=0\\ 2x=7\\ x=\boxed{\dfrac{7}{2}}$

Collecting terms, we have (2x -7)/[(x - 3)(x - 4)] = (2x -7)/{x _ 2)(x _ 5)]. The numerators are equal, the denominators are not. This can only happen if numerator = 0 = 2x -7, s- x = 3.5. Ed Gray

$\frac{1}{x-3}$ + $\frac{1}{x-4}$ = $\frac{1}{x-2}$ + $\frac{1}{x-5}$

$or$ , $\frac{1}{x-3}$ - $\frac{1}{x-2}$ = $\frac{1}{x-5}$ - $\frac{1}{x-4}$

$or,$ $\frac{x-3-x+2}{x^2-5x+6}$ = $\frac{x-5-x+4}{x^2-9x+20}$

$or,$ $-5x$ + $6$ = $-9x$ + $20$

$So$ , $x$ = $\frac{7}{2}$