Còn nha. Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}^{\left(1\right)}\)

Lại có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}^{\left(2\right)}\)

Từ (1) và (2) => đpcm

CMR:a²+b²/c²+d²=ab/cd=>a/b=c/d

Bài làm

a²+b²/c²+d²=ab/cd

=>(a²+b²)cd=>ab(c²+d²)

<=>a²(cd)+b²(cd)-abc²-abc²=0

<=>a²cd-abc²+b²cd-abc²=0

<=>ac(ad-bc)+bd(bc-ad)=0

<=>ac(ad-bc)-bd(bc-ad)=0

<=>(ac-bd)(ac-bd)=0

=>\(\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\)

=>\(\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\)

=>\(\orbr{\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}}\)=>ĐPCM

Từ \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\Rightarrow\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\left(1\right)\)

\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

Vậy    \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

bấm vào chỗ đúng đó nguyễn minh tâm

(a+b+c/b+c+d)^3=(a+b+c/b+c+d).(a+b+c/b+c+d).(a+b+c/b+c+d)=a/b.b/c.c/d

Ta có : \(b>0,d>0,\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)                                                                         ( 1 )

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

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

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

Vì \(b>0,d>0,\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow\frac{a}{b}< \frac{c}{d}=ad< bc\)

\(\Rightarrow ad+cd< bc+cd\)                                                             ( 2 )

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

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

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

+) Chứng minh: \(\frac{a}{b}>\frac{a+c}{b+d}\)   (1)

Xét hiệu: \(\frac{a}{b}-\frac{a+c}{b+d}=\frac{a\left(b+d\right)-b\left(a+c\right)}{b\left(b+d\right)}=\frac{ab+ad-ab-bc}{b\left(b+d\right)}=\frac{ad-bc}{b\left(b+d\right)}\)

Vì a/b > c/d ; b; d > 0 =>  ad > bc => ad - bc > 0 .T a có b(b +d) > 0 nên Hiệu trên > 0 => \(\frac{a}{b}>\frac{a+c}{b+d}\)

+) Chứng minh: \(\frac{a+c}{b+d}>\frac{c}{d}\)

Xét hiệu: \(\frac{a+c}{b+d}-\frac{c}{d}=\frac{\left(a+c\right)d-c\left(b+d\right)}{b\left(b+d\right)}=\frac{ad-bc}{b.\left(b+d\right)}>0\)

=> \(\frac{a+c}{b+d}>\frac{c}{d}\)  (2)

Từ (1)(2 ta có đpcm