a, \(\frac{x}{m}+1=m\)
=> \(\frac{x}{m}+\frac{m}{m}=\frac{m^2}{m}\)
=> \(x+m=m^2=>x+m-m^2=0\)
b, \(\frac{x-2m}{3}=8-\frac{m}{3}\)
=> \(\frac{x-2m}{3}=\frac{24}{3}-\frac{m}{3}\)
=> \(x-2m=24-m\)
=> \(x-24-m\)
c, \(\frac{x+a}{b}-\frac{b}{a}=\frac{x-b}{a}+\frac{a}{b}\)
=> \(\frac{a.\left(x+a\right)}{ab}-\frac{b^2}{ab}=\frac{b.\left(x-b\right)}{ab}+\frac{a^2}{ab}\)
=> \(ax+a^2-b^2=bx-b^2+a^2\)
=> \(ax-bx+\left(a^2-a^2\right)-\left(b^2-b^2\right)=0\)
=> \(x.\left(a-b\right)=0\)
=> x= 0 hoặc a - b =0