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Ta có :
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=\frac{ab-bc}{\left(a+b\right)-\left(b+c\right)}=\frac{bc-ca}{\left(b+c\right)-\left(c+a\right)}=\frac{ab-ca}{\left(a+b\right)-\left(c+a\right)}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow Q=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=1\)
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}\Leftrightarrow\left(a+b\right)\left(a-c\right)=\left(a+c\right)\left(a-b\right)\)
\(\Leftrightarrow a^2+ab-ac-bc=a^2+ac-ab-bc\Leftrightarrow ab-ac=ac-ab\)
<=>2ab=2ac<=>ab=ac<=>b=c
giờ thì dễ rồi, bạn tự thay vào biểu thức
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
=> \(\frac{2}{c}=\frac{a+b}{ab}\)
=> 2ab = ac + bc
=> ac + bc - 2ab = 0
=> (ac - ab) + (bc - ab) = 0
=> a(c - b) + b(c - a) = 0
=> a(c - b) = -b(c - a)
=> a(c - b) = b(a - c)
=> \(\frac{a}{b}=\frac{a-c}{c-b}\) (đpcm)
ta có:\(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{b}{a}=\frac{c}{d}\)
\(\Rightarrow1-\frac{b}{a}=1-\frac{c}{d}\)
\(\Rightarrow\frac{a}{a}-\frac{b}{a}=\frac{c}{c}-\frac{d}{c}\)
\(\Rightarrow\frac{a-b}{a}=\frac{c-d}{c}\)
hay: \(\frac{a}{a-b}=\frac{c}{c-d}\)(đpcm)
Cách 1 : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
Cách 2 : \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ac-ad=ac-bc\)
\(\Rightarrow a(c-d)=c(a-b)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
Cách 3 : Đặt \(\frac{a}{b}=\frac{c}{d}=m\Rightarrow a=mb,c=md\)
Ta có : \(\frac{a}{a-b}=\frac{mb}{mb-b}=\frac{mb}{b(m-1)}=\frac{m}{m-1}\)
\(\frac{c}{c-d}=\frac{md}{md-d}=\frac{md}{d(m-1)}=\frac{m}{m-1}\)
Do đó : \(\frac{a}{a-b}=\frac{c}{c-d}\)
Cách 4 : \(\frac{a}{a-b}=\frac{c}{c-d}\Rightarrow a(c-d)=c(a-b)\)
\(\Rightarrow ac-ad=ac-bc\Rightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\) đẳng thức đúng
Do đó , ta có : \(\frac{a}{a-b}=\frac{c}{c-d}\)là đẳng thức đúng.
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\left(1\right)\\ \Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\frac{a-b}{c-d}=\frac{ck-dk}{c-d}=\frac{k\left(c-d\right)}{c-d}=k\left(2\right)\)
(1)(2) \(\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{b}{a}=\frac{d}{c}\)
\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1\)
\(\Rightarrow\frac{b}{a}-\frac{a}{a}=\frac{d}{c}-\frac{c}{c}.\)
\(\Rightarrow\frac{b-a}{a}=\frac{d-c}{c}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\left(đpcm\right).\)
Chúc bạn học tốt!