Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

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

=>\(a^2cd+b^2cd-abc^2-abd^2=0\)

=>\(ac\left(ad-bc\right)+bd\left(bc-ad\right)=0\)

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

TH1: ac-bd=0

=>ac=bd

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

TH2: ad-bc=0

=>ad=bc

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

co ai biet ko? Neu biet thi giup mk voi

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

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

TH1: a-c\(\ne0\)<=>a+c=d+b<=>a-b=d-c cộng vế với vế với a+b=c+d (gt) <=> 2a=2d <=> a=d <=> b=c

=>a²⁰⁰⁶=d²⁰⁰⁶;b²⁰⁰⁶=c²⁰⁰⁶=>a²⁰⁰⁶+b²⁰⁰⁶=c²⁰⁰⁶+d²⁰⁰⁶

TH2: a-c=0 <=> a=c <=> b=d <=> a²⁰⁰⁶+b²⁰⁰⁶=c²⁰⁰⁶+d²⁰⁰⁶

Từ 2 trường hợp trên suy ra đpcm

Đặt\(\frac{a}{b}=\frac{c}{d}=k\left(k\in Q\right)\)

\(\Rightarrow\hept{\begin{cases}a=bk\left(1\right)\\c=dk\left(2\right)\end{cases}}\)

Ta lại có \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(3\right)\)

Thay \(\left(1\right),\left(2\right)vào\left(3\right)có\)

\(\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(4\right)\)

\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(5\right)\)

Từ \(\left(4\right),\left(5\right)\Rightarrowđpcm\)