\(a\)) Giải:
\(A=\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}=\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{5.\left(\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}=\frac{1}{5}\)
\(b\)) Giải:
\(B=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}=\frac{\left(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}\right).132}{\left(\frac{5}{12}+1-\frac{7}{11}\right).132}=\frac{88-33+60}{55+132-84}=\frac{115}{103}\)