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\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)
<=>\(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)
<=> \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)
<=> \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)
<=> \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)
<=> \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}\left(đpcm\right)}}\)
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Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
\(\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
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Đặt a/b=c/d= t suy ra a=bt; c=dt
(a+b)/(a-b)= bt+b/bt-b = b(t+1)/b(t-1)=t+1/t-1 (1)
(c+d)/(c-d)= dt+d/dt-d = d(t+1)/d(t-1)=t+1/t-1 (2)
Từ (1) và (2) suy ra (a+b)/(a-b)= (c+d)/(c-d)
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
\(\cdot\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)
\(\cdot\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\)\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
Chúc bạn học tốt!!! k cho mk nha !!
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\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
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Nếu ( a+b+c+d ) . ( a-b-c+d ) = ( a-b+c-d) . ( a+b-c-d)
=> \(\frac{a+b+c+d}{a-b+c-d}=\)\(\frac{a+b-c-d}{a-b-c+d}\)
=> \(\frac{a+b+c+d}{a-b+c-d}=\)\(\frac{a+b-c-d}{a-b-c+d}\)\(=\frac{\left(a+b+c+d\right)+\left(a+b-c-d\right)}{\left(a-b+c-d\right)+\left(a-b-c+d\right)}\)\(=\frac{2.\left(a+b\right)}{2.\left(a-b\right)}\)\(=\frac{a+b}{a-b}\)
và
\(\frac{a+b+c+d}{a-b+c-d}=\)\(\frac{a+b-c-d}{a-b-c+d}\)\(=\frac{\left(a+b+c+d\right)-\left(a+b-c-d\right)}{\left(a-b+c-d\right)-\left(a-b-c+d\right)}\)\(=\frac{2.\left(c+d\right)}{2.\left(c-d\right)}\)\(=\frac{c+d}{c-d}\)
=>\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
=>\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)\(=\frac{a+b+a-b}{c+d+c-d}=\frac{a+b-\left(a-b\right)}{c+d-\left(c-d\right)}\)=> \(\frac{2a}{2c}=\frac{2c}{2d}\)=> \(\frac{a}{c}=\frac{b}{d}\)hay \(\frac{a}{b}=\frac{c}{d}\)
Vậy \(\frac{a}{b}=\frac{c}{d}\)