dãy ti số bằng nhau 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó : \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\left(1\right)\)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2.\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2+d^2}\left(\text{đpcm}\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có

\(VT=\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\cdot k^2+b^2}{d^2\cdot k^2+d^2}=\frac{b^2\cdot\left(k^2+1\right)}{d^2\cdot\left(k^2+1\right)}=\frac{b^2}{d^2}\)

\(VT=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2\cdot k^2-b^2}{d^2\cdot k^2-d^2}=\frac{b^2\cdot\left(k^2-1\right)}{d^2\cdot\left(k^2-1\right)}=\frac{b^2}{d^2}\)

\(\Rightarrow VT=VP\)

\(\Leftrightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)

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Bài 1:Với  a,b,c,d dương

Ta có: \(\frac{a}{a+b+c+d}

đặt a+c vào 2bd ta có (a+c)d = c(b+d) <=> ad+ cd = bc + cd <=> ad = bc <=> a/ b = c/ d

(thay a+c vào 2bd vì a+c = 2b )

d(a+c)=2bd=c(b+d)

Suy ra ad+dc=cb+cd

ad=cb

Ta suy ra  được a/b=c/d

từ a+c= 2b 

=> d(a+c)=2bd

mà 2bd=c(b+d)

=> d(a+c) =c(b+d)

<=> ad +dc = cb +dc

<=>ad =cb 

=> a/b =c/d (đpcm)

Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{a-b+c-d}{b-c+d-e}\left(1\right)\)

Ta lại có: \(\left\{\begin{matrix}\frac{a}{b}=\frac{b}{c}\\\frac{c}{d}=\frac{d}{e}\\\frac{b}{c}=\frac{c}{d}\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}a=\frac{b^2}{c}\\e=\frac{d^2}{c}\\d=\frac{c^2}{b}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{e}=\frac{b^2}{d^2}\\d=\frac{c^2}{b}\end{matrix}\right.\)

\(\Rightarrow\frac{a}{e}=\frac{b^2}{\left(\frac{c^2}{b}\right)^2}=\frac{b^4}{c^4}\left(2\right)\)

Từ (1) và (2) suy ra: \(\frac{a}{e}=\left(\frac{a-b+c-d}{b-c+d-e}\right)^4\)