Question

Cho : \(\frac{a^2+b^2}{c^2+d^2}=\frac{a.b}{c.d}\)với \(a,b,c,d\ne0\)

CMR:\(\frac{a}{b}=\frac{c}{d}\)và \(\frac{a}{b}=\frac{d}{c}\)

 

Answer 1

dễ mk sao 0 biêy

Answer 2

Thế 

làm

đi ~.~

Answer 3

uk trong sgk cũng co

Answer 4

Lm đi !!!

T tham khảo cái

~.~

Answer 5

uk đợi chút

Answer 6

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

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

\(\Rightarrow cda^2+cdb^2=abc^2+abd^2\)

\(\Rightarrow cda^2+cdb^2-abc^2-abd^2=0\)

\(\Rightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)\)

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

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

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

Vậy...............

Answer 7

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

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

\(\Leftrightarrow a^2.cd+b^2.cd=abc^2+abd^2\)

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

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

\(\Leftrightarrow\left(ac-bd\right)\left(ac-bc\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}}\)( Đpcm )