a)$\frac{x-1}{x+2}$-$\frac{x}{x-2}$=$\frac{5x-2}{4-x^2}$ Đkxđ: x$\neq$±2
⇔$\frac{(x-1)(x-2)}{(x+2)(x-2)}$-$\frac{x(x+2)}{(x+2)(x-2)}$=-$\frac{5x-2}{(x+2)(x-2)}$
⇒(x-1)(x-2)-x(x+2)=-5x+2
⇔x²-3x+2-x²-2x=-5x+2
⇔0x=0 (luôn đúng)
Vậy S=R
b) $\frac{1}{x+1}$-$\frac{5}{x-2}$=$\frac{15}{(x+1)(2-x)}$ Đkxđ: x$\neq$-1; x$\neq$2
⇔$\frac{x-2}{(x+1)(x-2)}$-$\frac{5(x+1)}{(x+1)(x-2)}$=-$\frac{15}{(x+1)(x-2)}$
⇒x-2-5(x+1)=-15
⇔x-2-5x-5=-15
⇔-4x=-8
⇔x=2 (không thỏa mãn đkxđ)
Vậy phương trình đã cho vô nghiệm
c)$\frac{3x}{x-2}$-$\frac{x}{x-5}$=$\frac{3x}{(x-2)(5-x)}$ Đkxđ: x$\neq$2; x$\neq$5
⇔$\frac{3x(x-5)}{(x-2)(x-5)}$-$\frac{x(x-2)}{(x-2)(x-5)}$=-$\frac{3x}{(x-2)(x-5)}$
⇒3x(x-5)-x(x-2)=-3x
⇔3x²-15x-x²+2x=-3x
⇔2x²-10x=0
⇔2x(x-5)=0
⇔\(\left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=0.(tm.đkxđ)\\x=5.(không.tm.đkxđ)\end{array} \right.\)
Vậy S={0}.