\(\frac{x+2}{x-m}=\frac{x+1}{x-1}\) (ĐKXĐ: x \(\ne\) m)
\(\Leftrightarrow\) \(\frac{\left(x+2\right)\left(x-1\right)}{\left(x-m\right)\left(x-1\right)}=\frac{\left(x+1\right)\left(x-m\right)}{\left(x-m\right)\left(x-1\right)}\)
\(\Rightarrow\) (x + 2)(x - 1) = (x + 1)(x - m)
\(\Leftrightarrow\) x2 - x + 2x - 2 = x2 - mx + x - m
\(\Leftrightarrow\) x2 + x - 2 = x2 - (m-1)x - m
\(\Leftrightarrow\) x2 - x2 + x + (m-1)x + m = 2
\(\Leftrightarrow\) mx + m = 2
\(\Leftrightarrow\) m(x + 1) = 2
\(\Rightarrow\) m(x + 1) \(\in\) Ư(2)
Ư(2) = {1; 2; -1; -2}
\(\Rightarrow\) Nếu m = {1; 2; -1; -2}
\(\Rightarrow\) x + 1 = {2; 1; -2; -1}
hay x = {1; 0; -3; -2}
Vậy m \(\in\){1; 2; -1; -2} thì x có 1 nghiệm duy nhất
Ko bt đúng ko :v
