Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{a}{b}.bd< \frac{c}{d}.bd\Rightarrow ad< cb\)(đpcm)
b)Ta có:
- ad<cd
=>ab+ad<ab+cd
=>a(b+d)<b(b+d)
=>\(\frac{a\left(b+d\right)}{b\left(b+d\right)}< \frac{b\left(a+c\right)}{b\left(b+d\right)}\)
=>\(\frac{a}{b}< \frac{a+c}{b+d}\)(1)
- ad<bc
=>ad+cd<bc+cd
=>d(a+c)<c(b+d)
=>\(\frac{d\left(a+c\right)}{d\left(b+d\right)}< \frac{c\left(b+d\right)}{d\left(b+d\right)}\)
=>\(\frac{a+c}{b+d}< \frac{c}{d}\)(2)
Từ (1) và (2) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)(đpcm)
a) Ta có:
a/b < c/d
=> a/b . d/c < c/d . d/c
=> ad/bc < 1
=> ad < 1.bc
=> ad < bc ( đpcm)
b) Ta có: ad < bc
=> ad + ab < bc + ab
=> a.(b + d) < b.(a + c)
=> a/b < a+c/b+d (1)
Ta có: ad < bc
=> ad + cd < bc + cd
=> d.(a + c) < c.(b + d)
=> a+c/b+d < c/d (2)
Từ (1) và (2) => a/b < a+c/b+d < c/d ( đpcm)
Ta có:
\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\)
\(\Rightarrow\left\{{}\begin{matrix}ad+ab< bc+ab\\ad+cd< bc+cd\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a\left(b+d\right)< b\left(a+c\right)\\d\left(a+c\right)< c\left(b+d\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{a+c}{b+d}\\\dfrac{a+c}{b+d}< \dfrac{c}{d}\end{matrix}\right.\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{a}{d}\)(đpcm)
Chúc bạn học tốt!!!
a)\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow\frac{ad}{bd}< \frac{cb}{db}\Leftrightarrow ad< bc\left(đpcm\right)\)
b)
có\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+ab< bc+ab\)
\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Leftrightarrow\frac{a}{a+c}< \frac{b}{b+d}\)
\(\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)
có\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Leftrightarrow\frac{d}{b+d}< \frac{c}{a+c}\)
\(\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)
từ (1) và (2)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\left(đpcm\right)\)