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Đặt \(A=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
\(A=\frac{\frac{44}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)=\(\frac{\frac{44.1}{132}+-\frac{33.1}{132}+\frac{60.1}{132}}{\frac{55.1}{132}+\frac{132.1}{132}-\frac{84.1}{132}}\)
\(A=\frac{\frac{1}{132}.\left(44-33+60\right)}{\frac{1}{132}.\left(55+132-84\right)}\)=\(\frac{\frac{1}{132}.71}{\frac{1}{132}.103}\)
\(A=\frac{71}{103}\)
( \(\frac{2}{3}\)- \(\frac{1}{4}\)+ \(\frac{5}{11}\)) : ( \(\frac{5}{12}\)+ 1 - \(\frac{7}{11}\))
= ( \(\frac{5}{12}\)+ \(\frac{5}{11}\)) : ( \(\frac{5}{12}\)+ \(\frac{4}{11}\))
= \(\frac{115}{132}\): \(\frac{103}{132}\)
= \(\frac{115}{103}\)
(2/3 - 1/4 +5/11 ) : ( 5/12 +1 - 7/11 )
= ( 5/12 +5/11 ) : ( 17/12-7/11 )
= 115/132:103/132
=115/132*132/10
=115/103
Đặt A=2/3-1/4+5/11
A=5/12+5/11
A=115/132
Đặt B=5/12+1-7/11
B=103/132
Suy ra:P=A/B
P=115/132:103/132
P=115/103
A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12
\(P=\dfrac{\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{12}{12}-\dfrac{7}{11}}=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{17}{12}-\dfrac{7}{11}}=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{103}\)