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Đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)= k =>\(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có: \(\frac{a+b}{a-b}\)=\(\frac{bk+b}{bk-b}\)=\(\frac{b\left(k+1\right)}{b\left(k-1\right)}\)=\(\frac{k+1}{k-1}\)(1)
\(\frac{c+d}{c-d}\)=\(\frac{dk+d}{dk-d}\)=\(\frac{d\left(k+1\right)}{d\left(k-1\right)}\)=\(\frac{d+1}{d-1}\)(2)
Từ (1),(2) =>\(\frac{a+b}{a-b}\)=\(\frac{c+d}{c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}=\frac{a+b+c+d}{a+b+c+d}\)
Th1:a+b+c+d=0=>\(\frac{a+b+c+d}{a+b+c+d}=\frac{0}{a+b+c+d}=0suyra\frac{a+b}{b+c}=\frac{c+d}{d+a}=0\)
Th2:a+b+c+d khác 0=>\(\frac{a+b+c+d}{a+b+c+d}=1\)suy ra\(\frac{a+b}{b+a}=\frac{c+d}{d+a}=1\)=>(a+b)(d+a)=(b+a)(c+d)=>a+d=c+d<=>a=c
Vậy a+b+c+d=0 hoặc a=c
Ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)
\(\implies\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)
\(\implies\) \(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)
\(\implies\) \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)
\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)
\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)
\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)
\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)
\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c+d=d+a\end{cases}}\)
\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}}\)
`Answer:`
a. Ta đặt \(\hept{\begin{cases}k=\frac{a}{b}=\frac{c}{d}\\bk=a\\dk=c\end{cases}}\)
\(\Rightarrow\frac{a+b}{b}=\frac{b+bk}{b}=\frac{\left(k+1\right).b}{b}=k+1\left(1\right)\)
\(\Rightarrow\frac{c+d}{d}=\frac{d+dk}{d}=\frac{\left(k+1\right).d}{d}=k+1\left(2\right)\)
Từ `(1)(2)=>\frac{a+b}{b}=\frac{c+d}{d}`
Bạn tham khảo ở đây nha
http://olm.vn/hoi-dap/question/222914.html
chứng minh rằng từ tỉ lệ thức a/b=c/d(a khác b, c khác d)ta có thể suy ra tỉ lệ thức a+b/a-b=c+d/c-d
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Ta có:
\(VT=\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)
\(VP=\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)
Từ (1) và (2) ->đpcm
\(\frac{a}{b}=\frac{c}{d}\)(\(b,d\ne0\))
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow2ad=2bc\)
\(\Leftrightarrow ad-bc=bc-ad\)
\(\Leftrightarrow ad-bc+ac-bd=bc-ad+ac-bd\)
\(\Leftrightarrow\left(a+b\right)\left(c-d\right)=\left(c+d\right)\left(a-b\right)\)
\(\Leftrightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(\(a-b,c-d\ne0\))
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(a,\frac{a-b}{a}=\frac{c-d}{c}\)
\(VT=\frac{a-b}{a}=\frac{bk-b}{bk}=\frac{b\cdot\left(k-1\right)}{bk}=\frac{k-1}{k}\)
\(VP=\frac{c-d}{c}=\frac{dk-d}{dk}=\frac{d\cdot\left(k-1\right)}{dk}=\frac{k-1}{k}\)
\(\Rightarrow VT=VP\Leftrightarrow\frac{a-b}{a}=\frac{c-d}{c}\)
\(b,\frac{a-b}{a}=\frac{c+d}{c}\)
Bạn kt lại xem đúng chưa tớ nghĩ là \(\frac{a+b}{a}=\frac{c+d}{c}\)nên mik giải theo đề này
\(VT=\frac{a+b}{a}=\frac{bk+b}{bk}=\frac{b\cdot\left(k+1\right)}{bk}=\frac{k+1}{k}\)
\(VP=\frac{c+d}{c}=\frac{dk+d}{dk}=\frac{d\cdot\left(k+1\right)}{dk}=\frac{k+1}{k}\)
\(\Rightarrow VT=VP\Leftrightarrow\frac{a+b}{a}=\frac{c+d}{c}\)
a) Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\rightarrow\frac{c-d}{c}=\frac{a-b}{a}\)
b) Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)
\(\rightarrow\frac{c-d}{c}=\frac{a-b}{a}=\frac{a+b}{c+d}\)
\(\rightarrow\frac{a-b}{a}=\frac{c+d}{c}=\frac{a+b}{c-d}\rightarrow\frac{a-b}{a}=\frac{c+d}{c}\)