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![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a)
$\frac{a}{b}< \frac{c}{d}\Leftrightarrow \frac{ad}{bd}< \frac{bc}{bd}$
$\Leftrightarrow \frac{ad-bc}{bd}< 0$
Vì $bd>0$ với mọi $b,d>0$ nên $ad-bc< 0\Leftrightarrow ad< bc$
b) Từ phần a suy ra $bc-ad>0$
$\frac{a+c}{b+d}-\frac{a}{b}=\frac{b(a+c)-a(b+d)}{b(b+d)}=\frac{bc-ad}{b(b+d)}>0$ do $bc-ad>0$ và $b(b+d)>0$ với mọi $b,d>0$)
$\Rightarrow \frac{a+c}{b+d}>\frac{a}{b}$
Lại có:
$\frac{a+c}{b+d}-\frac{c}{d}=\frac{d(a+c)-c(b+d)}{d(b+d)}=\frac{ad-bc}{d(b+d)}<0$ do $ad-bc<0$ và $d(b+d)>0$ với mọi $b,d>0$
$\Rightarrow \frac{a+c}{b+d}< \frac{c}{d}$
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(a,b,c,d\in N^{\circledast}\) nên \(\left\{{}\begin{matrix}a+b+c< a+b+c+d\\a+b+d< a+b+c+d\\b+c+d< a+b+c+d\\a+c+d< a+b+c+d\end{matrix}\right.\)
Ta có :
\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\\ \dfrac{b}{a+b+d}>\dfrac{b}{a+b+c+d}\\ \dfrac{c}{b+c+d}>\dfrac{c}{a+b+c+d}\\ \dfrac{d}{a+c+d}>\dfrac{d}{a+b+c+d}\\ \Rightarrow P>\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}=1\\ \Rightarrow P>1\left(1\right)\)
Vì \(a,b,c,d\in N^{\circledast}\) nên \(\left\{{}\begin{matrix}a+b+c>d\\a+b+d>c\\b+c+d>a\\a+c+d>b\end{matrix}\right.\)
Ta có :
\(\dfrac{a}{a+b+c}=\dfrac{2a}{\left(a+b+c\right)+\left(a+b+c\right)}< \dfrac{2a}{a+b+c+d}\)
\(\dfrac{b}{a+b+d}=\dfrac{2b}{\left(a+b+d\right)+\left(a+b+d\right)}< \dfrac{2b}{a+b+c+d}\left(a+b+d>c\right)\\ \dfrac{c}{b+c+d}=\dfrac{2c}{\left(b+c+d\right)+\left(b+c+d\right)}< \dfrac{2c}{a+b+c+d}\left(b+c+d>a\right)\\ \dfrac{d}{a+c+d}=\dfrac{2d}{\left(a+c+d\right)+\left(a+c+d\right)}< \dfrac{2d}{a+b+c+d}\left(a+c+d>b\right)\)
Từ đó, ta có :
\(\dfrac{a}{a+b+d}+\dfrac{b}{a+b+d}+\dfrac{c}{b+c+d}+\dfrac{d}{a+c+d}< \\ \dfrac{2a}{a+b+c+d}+\dfrac{2b}{a+b+c+d}+\dfrac{2c}{a+b+c+d}+\dfrac{2d}{a+b+c+d}=2\\ \Rightarrow P< 2\left(2\right)\)
Từ (1) và (2), ta có điều phải chứng minh.
![](https://rs.olm.vn/images/avt/0.png?1311)
AD tích chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
\(\Rightarrow DPCM\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Vì $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$ nên:
$\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}$
Hay $\left(\frac{a}{b}\right)^3=\frac{a}{d}$
Ta có đpcm.
\(\left(\dfrac{a}{b}\right)^3=\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{a}{b}=\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{d}=\dfrac{a}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
Ta có:
Nếu:
\(\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\Leftrightarrow\left(2a+c\right)\left(b-d\right)=\left(a-c\right)\left(2b+d\right)\)
\(\Leftrightarrow2a\left(b-d\right)+c\left(b-d\right)=a\left(2b+d\right)-c\left(2b+d\right)\)
\(\Leftrightarrow2ab-2ad+bc-cd=2ab+ad-2bc+cd\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\left(đpcm\right)\)