a,

b,  a/b < c/d => ad < cb
=>ad +ab < bc+ab
=> a(d+b) < b(a+c)
=> a/b < a+c/d+b (1)
* a/b < c/d => ad<cb
=> ad + cd < cb +cd
=> d(a+c) < c(b+d) 
=> c/d > a+c/b+d (2)
Từ (1) và (2) => a/b < a+c/b+d < c/d

Vì \(b,d>0\)nên \(bd>0\)

Ta có:  \(\frac{a}{b}< \frac{c}{d}\)

\(\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\)

\(\Leftrightarrow ad< bc\)vì \(bd>0\)

(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)

=>\(\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)

=>\(\left(a+d\right)^2-\left(a-d\right)^2=\left(b+c\right)^2-\left(b-c\right)^2\)

=>(a+d-a+d)(a+d+a-d)=(b+c-b+c)(b+c+b-c)

=>\(2d\cdot2a=2c\cdot2b\)

=>ad=bc

=>\(\frac{a}{c}=\frac{b}{d}\)

\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< cb\) (1)

Ta quy đồng hai PS a/b và a+c/b+d để so sánh:

\(\frac{a}{b}...\frac{a+c}{b+d}\)

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

\(\Leftrightarrow ab+ad.....ab+cb\)

\(\Leftrightarrow ad....cb\)

Vì (1) => \(ad< cb\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(2\right)\)

Quy đồng PS a+c/b+d và c/d để so sánh ta được:

\(\frac{a+c}{b+d}....\frac{c}{d}\)

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

\(\Leftrightarrow ad+cd....+bc+cd\)

\(\Leftrightarrow ad...bc\)

Vì (1)

\(\Rightarrow ad< bc\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(3\right)\)

Từ (2) và (3) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\left(đpcm\right)\)

Chúc bạn học tốt !!!

Ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< cb\)

\(\Rightarrow ad+ab< bc+ab\)

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

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

Lại có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< cb\)

\(\Rightarrow ad+cd< cb+cd\)

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

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

Bài 1:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{5a+4b}{5a-4b}=\frac{5\cdot bk+4b}{5\cdot bk-4b}=\frac{b\left(5k+4\right)}{b\left(5k-4\right)}=\frac{5k+4}{5k-4}\)

\(\frac{5c+4d}{5c-4d}=\frac{5\cdot dk+4d}{5\cdot dk-4d}=\frac{d\left(5k+4\right)}{d\left(5k-4\right)}=\frac{5k+4}{5k-4}\)

Do đó: \(\frac{5a+4b}{5a-4b}=\frac{5c+4d}{5c-4d}\)

Bài 2:

a: Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>b=ck; a=bk=ck^2

\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(ck^2\right)^2+\left(ck\right)^2}{\left(ck\right)^2+c^2}=\frac{c^2k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=k^2\)

\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)

Do đó: \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

b: Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\)

Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\Leftrightarrow ad+ab< bc+ab\Leftrightarrow a\left(d+b\right)< b\left(c+a\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)(1)

\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow bc>ad\Leftrightarrow bc+cd>ad+cd\Leftrightarrow c\left(b+d\right)>d\left(a+c\right)\Leftrightarrow\frac{c}{d}>\frac{a+c}{b+d}\)(2)

Từ (1) và (2) suy ra điều phải chứng minh

Đề sai bạn ơi, đề đúng phải là nếu a+b/a-b = c+d/c-d thì a/b =c/d

Ta có: \(\left(a+b+c-d\right)\left(a-b-c-d\right)=\left(a+b-c+d\right)\left(a-b+c+d\right)\)

\(\Rightarrow\frac{a+b+c-d}{a+b-c+d}=\frac{a-b+c+d}{a-b-c-d}\Leftrightarrow\frac{\left(a+b\right)+\left(c-d\right)}{\left(a+b\right)-\left(c-d\right)}=\frac{\left(a-b\right)+\left(c+d\right)}{\left(a-b\right)-\left(c+d\right)}.\)

Đặt \(A=a+b;B=c-d;C=a-b;D=c+d.\)Ta được:

\(\frac{A+B}{A-B}=\frac{C+D}{C-D}\Rightarrow\frac{A}{B}=\frac{C}{D}\Leftrightarrow\frac{a+b}{c-d}=\frac{a-b}{c+d}\Rightarrow\frac{a+b}{a-b}=\frac{c-d}{c+d}\)

Vậy ta được:

\(\left(a+b+c-d\right)\left(a-b-c-d\right)=\left(a+b-c+d\right)\left(a-b+c+d\right)\)

\(\Rightarrow\frac{a+b}{a-b}=\frac{c-d}{c+d}.\)