2.
\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) . Ta có : +,ad < bc
\(\Rightarrow\)ad+ab < bc +ab (Cùng thêm ab vào 2 vế)
\(\Rightarrow\)a(b+d) < b(a+c)
\(\Rightarrow\)\(\dfrac{a}{b}\)< \(\dfrac{a+c}{b+d}\)
+, ad < bc
\(\Rightarrow\)ad + cd < bc + cd ( Cùng thêm cd vào 2 vế)
\(\Rightarrow\)d(a+c) < c(b+d)
\(\Rightarrow\)\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) Vậy \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
2.
ta có
\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\Rightarrow ad< bc\)
xét
\(\dfrac{a}{b}=\dfrac{a\left(b+d\right)}{b\left(b+d\right)}=\dfrac{ab+ad}{b\left(b+d\right)}\)
\(\dfrac{a+c}{b+d}=\dfrac{b\left(a+c\right)}{b\left(b+d\right)}=\dfrac{ab+bc}{b\left(b+d\right)}\)
vì \(\dfrac{ab+ad}{b\left(b+d\right)}< \dfrac{ab+bc}{b\left(b+d\right)}\left(ad< bc\right)\)
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)
xét
\(\dfrac{a+c}{b+d}=\dfrac{d\left(a+c\right)}{d\left(b+d\right)}=\dfrac{ad+cd}{d\left(b+d\right)}\)
\(\dfrac{c}{d}=\dfrac{c\left(b+d\right)}{d\left(b+d\right)}=\dfrac{bc+cd}{d\left(b+d\right)}\)
vì
\(\dfrac{ad+cd}{d\left(b+d\right)}< \dfrac{bc+cd}{d\left(b+d\right)}\left(ad< bc\right)\)
\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)
từ (1) và (2) => ĐPCM
Bài 1:
Để \(x\in Z\) => \(a-3\) \(⋮\) 2a
=> 2a - 6 \(⋮\) 2a
Mà 2a \(⋮\) 2a
=> 2a - 6 - 2a \(⋮\) 2a
=> -6 \(⋮\) 2a => -3 \(⋮\) a => \(a\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
Vậy a = -1;1;3;-3
Bài 2:
Vì \(\dfrac{a}{b}< \dfrac{c}{d}\) => \(ad< bc\)
+) \(ad< bc\) => \(ad+ab< bc+ab\)
=> \(a\left(b+d\right)< b\left(a+c\right)\)
=> \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (1)
+) \(ad< bc\) => \(ad+cd< bc+cd\)
=> \(d\left(a+c\right)< b\left(b+d\right)\)
=> \(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) (2)
Từ (1); (2) => đpcm
