Question

Cho a,b,c,d là các số thực thỏa mãn :\(\dfrac{a}{b}< \dfrac{c}{d}\)

so sánh: \(\dfrac{a}{b}\)với \(\dfrac{a+c}{b+d}\)

Answer 1

\(\dfrac{a}{b}< 1\Leftrightarrow a< b\) thì a+c < b+c

Ta có: \(\dfrac{a+c}{b+d}=1-\dfrac{b-a}{b+d}\)

\(\dfrac{a}{b}=\dfrac{b-a}{b}\)

Vì \(\dfrac{b-a}{b+d}< \dfrac{b-a}{b}\)

=> \(1-\dfrac{b-a}{b+d}>1-\dfrac{b-a}{b}\)

\(\Rightarrow\dfrac{a+c}{b+d}>\dfrac{a}{b}\).

Answer 2

sorry nha! mình bấm lộn rồi

\(\dfrac{a}{b}=1-\dfrac{b-a}{b}\)