Đáp án:
`S={0,\frac{1}{2}}`
Giải thích các bước giải:
` \frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}= \frac{2x+5}{x^2-4x+3} (ĐK: x \ne 1, x \ne 2, x \ne 3)` `<=> \frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}-\frac{2x+5}{x^2-4x+3}=0`
`<=> \frac{x+4}{x^2-x-2x+2}-\frac{x+1}{x^2-x-3x+3}- \frac{2x+5}{x^2-x-3x+3}=0`
`<=> \frac{x+4}{x(x-1)-2(x-1)}-\frac{x+1}{x(x-1)-3(x-1)}- \frac{2x+5}{x(x-1)-3(x-1)}=0`
`<=> \frac{x+4}{(x-1)(x-2)}-\frac{x+1}{(x-1)(x-3)}- \frac{2x+5}{(x-1)(x-3)}=0`
`<=> \frac{(x-3)(x+4)-(x-2)(x+1)-(x-2)(2x+5)}{(x-2)(x-1)(x-3)}=0`
`<=> \frac{x^2+4x-3x-12-(x^2+x-2x-2)-(2x^2+5x-4x-10)}{(x-2)(x-1)(x-3)}=0`
`<=> \frac{x^2+x-12-(x^2-x-2)-(2x^2+x-10)}{(x-2)(x-1)(x-3)}=0`
`<=> \frac{x^2+x-12-x^2+x+2-2x^2-x+10}{(x-2)(x-1)(x-3)}=0`
`<=> \frac{-2x^2+x}{(x-2)(x-1)(x-3)}=0`
`<=> -2x^2+x=0 <=> x(-2x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\-2x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=\frac{1}{2}\end{array} \right.\)`(tm)`
`KL:``S={0,\frac{1}{2}}`
