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\(\frac{a}{b}< \frac{c}{d}\)
\(\Rightarrow ad< bc\)
\(\Rightarrow ab+ad< bc+ab\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\)( 1 )
Lại có : ad < bc
\(\Rightarrow ad+cd< bc+cd\)
\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
a, \(\frac{a}{b}=\frac{ad}{bd};\frac{c}{d}=\frac{bc}{bd}\)
Mà \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Rightarrow ad< bc\)
b, Theo câu a ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\Rightarrow ad+ab< bc+ab\Rightarrow a\left(b+d\right)< b\left(a+c\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)
Lại có: \(ad< bc\Rightarrow ad+cd< bc+cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)
Từ (1) và (2) => đpcm
a, \(\frac{a}{b}=\frac{ad}{bd};\frac{c}{d}=\frac{bc}{bd}\)
Mà \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Rightarrow ad< bc\)
b, Theo câu a, ta có:
\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\Rightarrow ad+ab< bc+ab\Rightarrow a\left(b+d\right)< b\left(a+c\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\)(1)
Lại có: \(ad< bc\Rightarrow ad+cd< bc+cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\)(2)
Từ (1) và (2) => đpcm.
\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)
+) \(ad+ab< bc+ab\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)( 1 )
+) \(ad+cd< bc+cd\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\)
Vì \(b,d>0\Rightarrow bd>0\)
\(\Rightarrow ad< bc\)
Ta lại có:
\(\frac{a}{b}=\frac{a\left(b+d\right)}{b\left(b+d\right)}=\frac{ab+ad}{b\left(b+d\right)}\)
\(\frac{a+c}{b+d}=\frac{b\left(a+c\right)}{b\left(b+d\right)}=\frac{ab+bc}{b\left(b+d\right)}\)
Vì \(b,d>0\)
Nên \(b\left(b+d\right)>0\)và \(d\left(b+d\right)>0\) \(\left(1\right)\)
Mà \(ad< bc\Leftrightarrow ab+ad< ab+bc\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)ta có: \(\frac{ab+ad}{b\left(b+d\right)}>\frac{ab+bc}{b\left(b+d\right)}\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(\cdot\right)\)
Ta lại có:
\(\frac{a+c}{b+d}=\frac{d\left(a+c\right)}{d\left(b+d\right)}=\frac{ad+cd}{d\left(b+d\right)}\)
\(\frac{c}{d}=\frac{c\left(b+d\right)}{d\left(b+d\right)}=\frac{bc+cd}{d\left(b+d\right)}\)
Mà \(ad< bc\Rightarrow ad+cd< bc+cd\left(3\right)\)
Từ \(\left(1\right)\)và \(\left(3\right)\)ta có:
\(\frac{ad+cd}{d\left(b+d\right)}< \frac{bc+cd}{d\left(b+d\right)}\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(\cdot\cdot\right)\)
Từ \(\left(\cdot\right)\)và \(\left(\cdot\cdot\right)\)ta có: \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
\(\frac{a}{b}<\frac{c}{d}\Rightarrow\)ad < bc. (1)
Từ (1) \(\Rightarrow\frac{a.\left(b+d\right)}{b.\left(b+d\right)}<\frac{\left(a+c\right).b}{\left(b+d\right).b}\Leftrightarrow\frac{a}{b}<\frac{a+c}{b+d}\) (2)
Từ (1) cũng \(\Rightarrow\frac{\left(a+c\right).d}{\left(b+d\right).d}<\frac{c.\left(b+d\right)}{d.\left(b+d\right)}\Leftrightarrow\frac{a+c}{b+d}<\frac{c}{d}\) (3)
Từ (2) và (3) suy ra điều phải chứng minh.
Vì \(\frac{a}{b}\) < \(\frac{c}{d}\) nên ad < bc (1)
Xét tích a(b + d) = ab + ad (2)
b ( a + c ) = ba + bc (3)
Từ (1);(2);(3) suy ra a(b+d) < b(a+c) do đó \(\frac{a}{b}\) < \(\frac{a+c}{b+d}\) (4)
Tương tự ta có \(\frac{a+c}{b+d}\) < \(\frac{c}{d}\) (5)
kết hợp (4) ; (5) ta được \(\frac{a}{b}\) < \(\frac{a+c}{b+d}\) < \(\frac{c}{d}\)
vì \(\frac{a}{b}< \frac{c}{d}=>ad< bc\)
=>ad+ab<bc+ab
=>a(b+d)<b(a+c)
=>\(\frac{a}{b}< \frac{a+c}{b+d}\) (1)
vì \(\frac{a}{b}< \frac{c}{d}=>ad< bc\)
=>ad+cd<bc+cd
=>a(a+c)<c(b+d)
=>\(\frac{a+c}{b+d}< \frac{c}{d}\) (2)
từ (1)(2)=>\(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
chúc bạn học tốt
\(\frac{a}{b}< \frac{c}{d}\)
\(\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+ab< bc+ab\)
\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
\(\frac{a}{b}< \frac{c}{d}\)
\(\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)