Chọn: Đáp án B

Ta có: ${{\log }_{140}}63={{\log }_{140}}\left( {{3}^{2}}.7 \right)=2{{\log }_{140}}3+{{\log }_{140}}7(1)$

${{\log }_{140}}3=\frac{1}{{{\log }_{3}}140}=\frac{1}{{{\log }_{3}}\left( {{2}^{2}}.5.7 \right)}=\frac{1}{2{{\log }_{3}}2+{{\log }_{3}}5+{{\log }_{3}}7}$

$=\frac{1}{2{{\log }_{3}}2+{{\log }_{3}}5+{{\log }_{3}}2.{{\log }_{2}}7}=\frac{1}{2.\frac{1}{a}+b+\frac{1}{a}.\frac{1}{c}}=\frac{ac}{2x+abc+1}$

${{\log }_{140}}7=\frac{1}{{{\log }_{7}}140}=\frac{1}{{{\log }_{7}}\left( {{2}^{2}}.5.7 \right)}=\frac{1}{2{{\log }_{7}}2+{{\log }_{7}}5+1}$

$=\frac{1}{2{{\log }_{7}}2+{{\log }_{7}}2.{{\log }_{2}}3.{{\log }_{3}}5+1}=\frac{1}{2c+abc+1}$

Từ đó suy ra: ${{\log }_{140}}63=2.\frac{ac}{2c+abc+1}+\frac{1}{2c+abc+1}=\frac{2ac+1}{2c+abc+1}$