Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?

ta có \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\Rightarrow ab.\left(c^2+d^2\right)=cd.\left(a^2+b^2\right)\)

suy ra \(ab.\left(c^2+d^2\right)\)=\(abc^2+abd^2=acbc+adbd\) (1)

\(cd\left(a^2+b^2\right)=a^2cd+b^2cd+bcbd\) =acad+bcbd (2)

(1);(2) suy ra acbc+adbd=acad+bcbd

nên bc+ad=bc+ad

suy ra ad=bc nên \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)

Ta đặt : a/b = c/d = K ( K khác 0 )

=> a = b.K

c = d.K

Mà : a^{2 +} b² / c² + d² = b.K² + b² / d.K² + d²

= b² . ( K² + 1 ) / d² . ( K² + 1 )

= b² / d² ( 1 )

Mà : ab/cd = b.K.b / d.K.d = b² . K / d² . K

= b² / d² ( 2 )

Từ ( 1 ) và ( 2 ) suy ra : a/b =c/d ( ĐPCM )

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\\ \Rightarrow\left(a^2+b^2\right).cd=ab\left(c^2+d^2\right)\\ \\ \Rightarrow a^2cd+b^2cd=abc^2+abd^2\\ \Rightarrow adac+bcbd=acbc+adbd\\ \Rightarrow ac\left(ad-bc\right)=bd\left(ad-bc\right)\\ \Rightarrow\left(ad-bc\right)\left(ac-bd\right)=0\\ \Rightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{c}{d}\\\dfrac{a}{b}=\dfrac{d}{c}\end{matrix}\right.\)

Biết   \(\dfrac{a^2 + b^2}{c^2 + d^2}=\dfrac{ab}{cd}\) với a,b,c,d khác 0. Chứng minh rằng:

\(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc\(\dfrac{a}{b}=\dfrac{d}{c}\) cái \(\dfrac{a}{b}=\dfrac{c}{d}\)thì mình chứng minh được rồi còn cái\(\dfrac{a}{b}=\dfrac{d}{c}\)thì chưa mong các bạn giúp ạ

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

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

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

\(\Leftrightarrow\left(a^2cd+b^2cd\right)-\left(c^2ab+d^2ab\right)=0\)

\(\Leftrightarrow aacd+bbcd-ccab-ddab=0\)(tất cả là dấu nhân ko phải số tự nhiên có 4 chữ số nha)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}ad-bc=0\\ac-bd=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\left(\text{đ}pcm\right)\)