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1
$m＜\frac{4}{3}$且$m≠\frac{2}{3}$
12.5
$\frac{4mn}{m-n}$
解：原式​$=\frac {1}{x-1}+\frac {x-1}{x+2}÷\frac {(x-1)^2}{(x+2)(x-2)}$​
​$ =\frac {1}{x-1}+\frac {x-1}{x+2}·\frac {(x+2)(x-2)}{(x-1)^2}$​
​$ =\frac {1}{x-1}+\frac {x-2}{x-1}$​
​$ =1$​
解：原式​$=(\frac {4m+5}{m+1}+\frac {\mathrm {m^2}-1}{m+1})·\frac {m+1}{m+2}$​
​$ =\frac {(m+2)^2}{m+1}·\frac {m+1}{m+2}$​
​$ =m+2$​
解：方程两边同乘​$x(2x-5)$​，得
​$ x=3(2x-5)$​
​$ $​解得​$x=3$​
检验：当​$x=3$​时，​$x(2x-5)≠0$​
∴该分式方程的解为​$x=3$​
解：方程两边同乘​$(x-1)(x+2)$​，得
​$ x^2+2x-3=(x-1)(x+2)$​
​$ $​解得​$x=1$​
检验：当​$x=1$​时，​$(x-1)(x+2)=0$​
∴​$x=1$​是原方程的增根，即原方程无解
③
$\frac{1-x}{x-2}$
解：选择​$(\boldsymbol {②}-\boldsymbol {①})×\boldsymbol {③}$​：
​$ (\boldsymbol {②}-\boldsymbol {①})×\boldsymbol {③}=(\frac {x-2}{x-1}-\frac {x^2-4x+4}{x^2-1})×\frac {2-2x}{2x-4}$​
​$ =[\frac {(x-2)(x+1)}{(x+1)(x-1)}-\frac {x^2-4x+4}{(x+1)(x-1)}]×\frac {1-x}{x-2}$​
​$ =\frac {3(x-2)}{(x+1)(x-1)}×\frac {1-x}{x-2}$​
​$ =-\frac {3}{x+1}$​
​$ $​当​$x=-4$​时，原式​$=-\frac {3}{-4+1}=1$​
​$ ($​或选择​$\boldsymbol {②}÷\boldsymbol {①}+\boldsymbol {③}$​：
​$ \boldsymbol {②}÷\boldsymbol {①}+\boldsymbol {③}=\frac {x-2}{x-1}÷\frac {x^2-4x+4}{x^2-1}+\frac {2-2x}{2x-4}$​
​$ =\frac {x-2}{x-1}×\frac {(x+1)(x-1)}{(x-2)^2}+\frac {1-x}{x-2}$​
​$ =\frac {x+1}{x-2}+\frac {1-x}{x-2}$​
​$ =\frac {2}{x-2}$​
​$ $​当​$x=-4$​时，原式​$=\frac {2}{-4-2}=-\frac {1}{3})$​
$\frac{12}{8^2-4}×(2-\frac{6-4}{6})=\frac{2}{6}$
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