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Ta có :
\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}-\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ ( 1 ) và ( 2 ) suy ra : \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{b}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\text{ hay }\frac{a}{b}=\frac{c}{d}\)
TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)
Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\text{ hay }\frac{a}{b}=\frac{d}{c}\)
Vậy : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\text{ thì }\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
kinh quá
Ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)
\(\implies\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)
\(\implies\) \(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)
\(\implies\) \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)
\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)
\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)
\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)
\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)
\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c+d=d+a\end{cases}}\)
\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}}\)
ta có \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)
=>\(\left(a+b\right)\left(a+d\right)=\left(c+d\right)\left(b+c\right)\)
=> \(a^2+ab+ad+bd=c^2+bc+bd+cd\)
=>\(a^2+ab+ad-bc-c^2-cd=0\)
=>\(\left(a^2-c^2\right)+\left(ab-cd\right)+\left(ab-ac\right)=0\)
=>\(\left(a-c\right)\left(a+c\right)+d\left(a-c\right)+b\left(a-c\right)=0\)
=>\(\left(a-c\right)\left(a+b+c+d\right)=0\)
=>\(\orbr{\begin{cases}a-c=0\\a+b+c+d=0\end{cases}\left(dpcm\right)}\)
hacker 2k6
Ta có: 2bd = c(b + d)
=> (a + c).d = bc + cd
=> ad + cd = bc + cd
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
Ta có : 2bd = c (b + d )
=) ( a + c ). d = bc + cd
=) ad + cd = bc + cd
=) ad = bc
=) a/b = c/ d ( đpcm)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Theo tính chất dãy tỉ số bằng nhau có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
ta có a+b/a-b=c+d/c-d
suy ra (a+b)(c-d)=(a-b)(c+d)
ac-ad+bc-bd=ac+ad-bc-bd
ac-ac+bc+bc-bd+bd=ad+ad
2bc=2ad
nen bc=ad=a/b=c/d
vay tu a/b=c/d ta co the suy ra a+b/a-b=c+d/c-d
\(\frac{a+b}{b+c}=\frac{c+d}{d+a}=\frac{a+b+c+d}{b+c+d+a}=1\) (dãy tỉ số bằng nhau)
\(\Rightarrow\frac{a+b}{a+c}=1\Leftrightarrow a+b=b+c\Rightarrow a=c\)(đpcm)
Từ : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\).
Theo tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{a+b}{b+c}=\frac{c+d}{d+a}=\frac{a+b-b-c}{a+d-d-a}=\frac{a-c}{c-a}\)
Nếu \(a-c=0\) thì \(a=c\)
Nếu : \(a-c\ne0\) thì \(\frac{a+b}{c+d}=-1\Rightarrow a+b=-c-d\Rightarrow a+b+c+d=0\)
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 \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)