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Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2+b^2}{ab}=\frac{c^2+d^2}{cd}\)
=> \(\frac{a^2}{ab}+\frac{b^2}{ab}=\frac{c^2}{cd}+\frac{d^2}{cd}\)
=> \(\frac{a}{b}+\frac{b}{a}=\frac{c}{d}+\frac{d}{c}\)
Mình chỉ làm được tới khúc này
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) suy ra:
\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
Trường hợp 1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) suy ra \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
Trường hợp 2: \(\frac{a+b}{c+d}=\frac{-\left(a-b\right)}{c-d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)
Từ (5) và (6) suy ra \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Rightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd-abc^2-abc^2=0\)
\(\Leftrightarrow a^2cd-abc^2+b^2cd-abc^2=0\)
\(\Leftrightarrow ac\left(ad-bc\right)+bd\left(bc-ad\right)=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ad-bc\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\Rightarrowđpcm\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
=> \(\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd-abc^2-abd^2=0\)
\(\Leftrightarrow a^2cd-abc^2+b^2cd-abd^2=0\)
\(\Leftrightarrow ac\left(ad-bc\right)+bd\left(bc-ad\right)=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ad-bc\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (DPCM)
Ta có \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+2ab+b^2}{c^2+2cd+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2-2ab+b^2}{c^2-2cd+d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\left(2\right)\)
Từ điều (1) và (2)
\(\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\left(a+b\right)\left(c-d\right)=\left(a-b\right)\left(c+d\right)\)
\(\Rightarrow c\left(a+b\right)-d\left(a+b\right)=c\left(a-b\right)+d\left(a-b\right)\)
\(\Rightarrow ac+bc-ad-bd=ac-bc+ad-bd\)
\(\Rightarrow bc-ad=-bc+ad\)
\(\Rightarrow2bc=2ad\)
\(\Rightarrow bc=ad\)
\(\Rightarrow\left[\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{b}{a}=\frac{d}{c}\end{matrix}\right.\) ( đpcm )
đề sai phải là CMR \(\frac{a}{b}=\frac{c}{d}\) hoặc \(\frac{b}{a}=\frac{d}{c}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Rightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{d}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\).
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) => \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
TH2: \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b+b-a}{c+d+d-c}=\frac{2b}{2d}=\frac{b}{d}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b-b+a}{c+d-d+c}=\frac{2a}{2c}=\frac{a}{c}\left(6\right)\)
Từ (5) và (6) => \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
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Ta có :
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\)( 1 )
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\)( 3 )
TH2 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\)( 4 )
Từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\)hay \(\frac{a}{b}=\frac{c}{d}\)
TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{2b}{2c}=\frac{b}{c}\)( 5 )
\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{2a}{2d}=\frac{a}{d}\)( 6 )
Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\)hay \(\frac{a}{b}=\frac{d}{c}\)
Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)