\(a,ad-bc=1\\ =>ad>bc\\ =>\dfrac{a}{b}>\dfrac{c}{d}\left(1\right)\\ cn-dm=1\\ =>cn>dm\\ =>\dfrac{c}{d}>\dfrac{m}{n}\left(2\right)\left(vìb,d,n>0\right)\)
Từ (1) và (2) suy ra \(\dfrac{a}{b}>\dfrac{c}{d}>\dfrac{m}{n}.Vậyx>y>z\)
\(b,ad-bc=cn-dm=1\\ =>ad+dm=bc+cn;d\left(a+m\right)=c\left(b+n\right)\)
Vậy \(\dfrac{c}{d}=\dfrac{a+m}{b+n}=>y=t\)