`a/b<(a+c)/(b+d)`

`<=>a(b+d)<b(a+c)`

`<=>ab+ad<ad<bc`

`<=>ad<bc`

`<=>a/b<c/d`(theo giả thiết)

`(a+c)/(b+d)<c/d`

`<=>d(a+c)<c(b+d)`

`<=>ad+cd<bc+dc`

`<=>ad<bc`

`<=>a/b<c/d`(theo giả thiết)`

`=>a/b<(a+c)/(b+d)<c/d`

Lời giải:
a. 

$\frac{a}{b}< \frac{c}{d}\Rightarrow \frac{a}{b}-\frac{c}{d}<0$

$\Rightarrow \frac{ad-bc}{bd}< 0$

$\Rightarrow ad-bc<0$ (do $bd>0$)

$\Rightarrow ad< bc$ (đpcm)

b.

$\frac{a}{b}-\frac{a+c}{b+d}=\frac{a(b+d)-b(a+c)}{b(b+d)}=\frac{ad-bc}{b(b+d)}<0$ do $ad-bc<0$ và $b(b+d)>0$

$\Rightarrow \frac{a}{b}< \frac{a+c}{b+d}$

--------

$\frac{a+c}{b+d}-\frac{c}{d}=\frac{d(a+c)-c(b+d)}{d(b+d)}=\frac{ad-bc}{d(b+d)}<0$ do $ad-bc<0$ và $d(b+d)>0$

$\Rightarrow \frac{a+c}{b+d}< \frac{c}{d}$
Ta có đpcm.

a) Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\)

\(\Rightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\)

\(\Rightarrow ad< bc\) ( đpcm. )

b) Vì \(b>0;d>0\) \(\Rightarrow b+d>0\)

Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\)

\(\Leftrightarrow ad< bc\) (*)

Thêm \(ab\) vào \(2\) vế (*), ta có:

\(ab+ad< ba+bc\)

\(a.\left(b+d\right)< b.\left(a+c\right)\)

\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)

Thêm \(cd\) vào \(2\) vế (*), ta được:

\(ad+cd< cb+cd\)

\(\left(a+c\right).d< c.\left(b+d\right)\)

\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:

\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\) ( đpcm )

a)ta có \(\dfrac{a}{b}\)<\(\dfrac{c}{d}\)\(\Rightarrow\)\(\dfrac{a\times d}{b\times d}\)=\(\dfrac{c\times b}{d\times b}\)\(\Rightarrow\)a\(\times\)d=c\(\times\)d\(\Rightarrow\)ad=bc

b)theo câu a ta có \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad=bc\)(1)

Thêm ab vào 2 vế của (1):ad+ab=bc+ab

a(b+d)<b(a+c)\(\Rightarrow\)\(\dfrac{a}{b}< \dfrac{a+c}{b+d}\)(2)

Thêm cd vào 2 vế của (1):ad+cd<bc+cd

d(a+c)<c(b+d)\(\Rightarrow\)\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\)(3)

Từ(2)và(3)\(\Rightarrow\)\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

Câu 1:

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\left(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}\right)\) - \(\left(\dfrac{x+1}{13}+\dfrac{x+1}{14}\right)=0\)

\(\Rightarrow\left(x+1\right).\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)\)= 0

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)

\(\Rightarrow x+1=0\)

=> x = 0 - 1

=> x = -1

Câu 2:

Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-3.4+9+12}{n-4}\)

\(=\dfrac{3.\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)

Để A có giá trị nguyên thì:

n - 4 \(\in\) Ư(21)

=> n - 4 \(\in\)

Câu 3:

Xét: \(\dfrac{a}{b}-\dfrac{a+c}{b+c}=\dfrac{a\left(b+d\right)-\left(a+c\right).b}{b\left(b+d\right)}\)

\(=\dfrac{ab+ad-ab-cb}{b\left(b+d\right)}=\dfrac{ad-cb}{b\left(b+d\right)}\)

ta thấy : ad<cb

=> ad - cb < 0 hay \(\dfrac{a}{b}-\dfrac{a+c}{b+d}< 0\)

=> \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\)

Hoàn toàn tương tự ta chứng minh được:

\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\)

`a)a/b<c/d`
Nhân 2 vế cho `bd>0` ta có:
`(abd)/b<(bcd)/d`
`<=>ad<bc`
`b)ad<bc`
Chia 2 vế cho `bd>0` ta có:
`(ad)/(bd)<(bc)/(bd)`
`<=>a/b<c/d`.

Đặt \(\left\{{}\begin{matrix}\dfrac{a}{b^2}=x\\\dfrac{b}{c^2}=y\\\dfrac{c}{a^2}=z\end{matrix}\right.\Rightarrow xyz=1;x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

Ta có \(x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\(\Leftrightarrow x+y+z=xy+yz+zx\)

\(\Leftrightarrow xyz-1+x+y+z-xy-yz-zx=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)

​\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)​

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b^2}=1\\\dfrac{b}{c^2}=1\\\dfrac{c}{a^2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b^2\\b=c^2\\c=a^2\end{matrix}\right.\left(đpcm\right)\)

Hằng đẳng thức:

\(\left(x-y-z\right)^2=x^2+y^2+z^2+2\left(yz-xy-zx\right)=x^2+y^2+z^2-2\left(xy+xz-yz\right)\)

\(\Rightarrow x^2+y^2+z^2=\left(x-y-z\right)^2+2\left(xy+xz-yz\right)\)

Giờ thay \(x=\dfrac{1}{a}\) ; \(y=\dfrac{1}{b}\); \(z=\dfrac{1}{c}\) là ra cái người ta làm