Ta có:
\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)=>ad+ab<bc+ab
=>a(b+d)>b(a+c)
=>\(\frac{a}{b}< \frac{a+c}{b+d}\) (1)
\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)=>ad+cd<bc+cd
=>d(a+c)<c(b+d)
=>\(\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) và (2) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)(đpcm)
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\(\frac{-1}{3}=\frac{-8}{24}>\frac{-9}{24}>\frac{-10}{24}>\frac{-11}{24}>\frac{-12}{24}=\frac{-1}{2}\)
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\(\frac{-1}{5}< \frac{-1}{4}< \frac{-1}{3}< \frac{-1}{2}< -1< 0< \frac{1}{5}\)
\(\frac{-1}{2}=\frac{\left(-1\right).12}{2.12}=\frac{-12}{24}\)
\(\frac{-1}{3}=\frac{\left(-1\right).8}{3.8}=\frac{-8}{24}\)
\(\frac{-8}{24}< x< \frac{-12}{24}\)
\(\Rightarrow x=\left\{\frac{-9}{24};\frac{-10}{24};\frac{-11}{24}\right\}\)
-4/2>-4/7>-2/3>-4/5>-4/3
-3/5<-2/5<-1/5<0/5<1/5<2/5<3/5
\(\frac{-1}{5}=\frac{-3}{15}\)
\(\frac{1}{5}=\frac{3}{15}\)
\(\frac{-3}{15}< x< \frac{3}{15}\)
\(\Rightarrow x=\left\{\frac{-2}{15};\frac{-1}{15};\frac{0}{15};\frac{1}{15};\frac{2}{15}\right\}\)
a/b<c/d =>a x d < b x c =>a x d+a x b < b x c+a x b =>a x (b+d) < b x (a+c) =>a/b < a+c/b+d
=>a x d < b x c =>a x d+ d x c< b x c+d x c =>d x (a+c) < c x (b+d) =>a+c/b+d < c/d
