Đáp án: $\frac{6x+1}{x²-7x+10}$ + $\frac{5}{x-2}$ = $\frac{3}{x-5}$ đkxđ: x$\neq$ 2 và 5
⇔ $\frac{6x+1}{x²-2x-5x+10}$ + $\frac{5}{x-2}$ = $\frac{3}{x-5}$
⇔ $\frac{6x+1}{(x²-2x)+(-5x+10)}$ + $\frac{5}{x-2}$ = $\frac{3}{x-5}$
⇔ $\frac{6x+1}{x(x-2)-5(x-2)}$ + $\frac{5}{x-2}$ = $\frac{3}{x-5}$
⇔ $\frac{6x+1}{(x-2)(x-5)}$ + $\frac{5}{x-2}$ = $\frac{3}{x-5}$
⇔ $\frac{6x+1}{(x-2)(x-5)}$ + $\frac{5.(x-5)}{(x-2)(x-5)}$ = $\frac{3.(x-2)}{(x-5)(x-2)}$
⇒ 6x+1+5x-25=3x-6
⇔6x+5x-3x = -6-1+25
⇔ 8x = 18
⇔ x = $\frac{18}{8}$
⇔ x = $\frac{9}{4}$
