Question

Chứng minh rằng:

Nếu \(\dfrac{a^2+b^2}{c^2+d^2}\) = \(\dfrac{ab}{cd}\) thì \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)

Answer 1

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 )

Answer 2

Answer 3

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