\(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}=\frac{\frac{3}{2}}{3a-\frac{3}{2}}=\frac{2.\frac{2}{3}}{2b-\frac{2}{3}}=\frac{3.\frac{1}{4}}{c-\frac{1}{4}}=\)
\(\frac{\frac{3}{2}}{3a-\frac{3}{2}}=\frac{\frac{4}{3}}{2b-\frac{2}{3}}=\frac{\frac{3}{4}}{c-\frac{1}{4}}=\frac{\frac{3}{2}+\frac{4}{3}-\frac{3}{4}}{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}=\frac{\frac{25}{12}}{4-\frac{23}{12}}=\frac{\frac{25}{12}}{\frac{25}{12}}=1\)
\(\Rightarrow\frac{1}{2a-1}=1\Rightarrow1=2a-1\Rightarrow a=1\)
Tương tự với b và c
