ta có:\(\dfrac{a}{b}< \dfrac{c}{d}=>a.d< c.b\)
ad+ab<cb+ab
hay a.(d+b)<b.(c+a)
=>\(\dfrac{a}{b}< \dfrac{c+a}{d+b}\)(1)
ad<cb
=>ad+dc<bc+cd
d.(a+c)<c.(b+d)
=>\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\)(2)
từ (1) và (2) ta có :
=>\(\dfrac{a}{b}< \dfrac{c+a}{d+b}\)\(< \dfrac{c}{d}\)
Tick đi ahihi :D