a: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)

hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)

\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)

Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{e}{f}=\dfrac{a+b+c+d+e}{b+c+d+e+f}=k\)

Ta có:

\(\dfrac{a}{f}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}.\dfrac{e}{f}=k^5=\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5\)

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)

Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)

DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?

Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)

\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

Vì \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

=> \(\dfrac{\left(a+b\right)^{2014}}{\left(c+d\right)^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}\)

Mà \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

=> \(\dfrac{\left(a+b\right)^{2014}}{\left(c+d\right)^{2014}}=\dfrac{\left(a-b\right)^{2014}}{\left(c-d\right)^{2014}}=\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}\) (1)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\) (2)

Từ (1);(2) => \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)