P=\(\frac{\frac{44}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}=\frac{\frac{44\cdot1}{132}+\frac{-33\cdot1}{132}+\frac{60\cdot1}{132}}{\frac{55\cdot1}{132}+\frac{132\cdot1}{132}-\frac{84\cdot1}{132}}\)
\(P=\frac{\frac{1}{132}\left(44-33+60\right)}{\frac{1}{132}\left(55+132-84\right)}=\frac{\frac{1}{132}\cdot71}{\frac{1}{132}\cdot103}\)
\(P=\frac{71}{103}\)
\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{6}{11}}=1\)
ta có:\(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}=\frac{88}{132}-\frac{33}{132}+\frac{60}{132}=\frac{88-33+60}{132}=\frac{115}{132}\)(1)
\(\frac{5}{12}+1-\frac{6}{11}=\frac{55}{132}+\frac{132}{132}-\frac{72}{132}=\frac{115}{132}\)(2)
từ (1)(2) suy ra \(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{6}{11}}=1\)