\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b^2\right)}{\left(c+d\right)^2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\Rightarrow\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\left(dpcm\right)\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{2b}{2d}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{2b}{2d}=\frac{a-2b}{c-2d}\)

\(\Rightarrow\frac{a^2}{c^2}=\frac{\left(a-2b\right)^2}{\left(c-2d\right)^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)(vì \(\frac{a}{c}=\frac{b}{d}\))

\(\Rightarrow\frac{ab}{cd}=\frac{\left(a-2b\right)^2}{\left(c-2d\right)^2}\left(đpcm\right)\)

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\)

\(\Rightarrow\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)

Vậy ...

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

Khi đó, ta có : \(\frac{bk.dk}{bd}=\frac{bdk^2}{bd}=k^2\)(1)

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

Từ (1) và (2) suy ra : \(\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)

\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\left(đpcm\right)\)

Đặt : \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Khi đó : \(\frac{\left(bk\right)^2-b^2}{kb^2}=\frac{\left(dk\right)^2-d^2}{kd^2}\)

\(\Rightarrow\frac{b^2.k^2-b^2}{kb^2}=\frac{d^2.k^2-d^2}{kd^2}\)

\(\Rightarrow\frac{b^2\left(k^2-1\right)}{kb^2}=\frac{d^2\left(k^2-1\right)}{kd^2}\)

\(\Rightarrow\frac{k^2-1}{k}=\frac{k^2-1}{k}\left(đpcm\right)\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^{1994}}{b^{1994}}=\frac{c^{1994}}{d^{1994}}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\frac{a^{1994}}{b^{1994}}=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}\)(1)

\(\frac{a^{1994}}{b^{1994}}=\frac{c^{1994}}{d^{1994}}=\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}\)(2)

từ (1) và (2) => \(\frac{a^{1994}+c^{1994}}{b^{1994}+d^{1994}}=\frac{\left(a+c\right)^{1994}}{\left(b+d\right)^{1994}}\left(đpcm\right)\)

\(\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{3a^6+c^6}{3b^6+d^6}\left(1\right)\)

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\left(2\right)\)

từ (1) và (2) => đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a=bk,c=dk

Ta có: \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{\left(bk\right)^2+b^2}=\frac{\left[b\left(k+1\right)\right]^2}{b^2k^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\left(1\right)\)

\(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(dk+d\right)^2}{\left(dk\right)^2+d^2}=\frac{\left[d\left(k+1\right)\right]^2}{d^2k^2+d^2}=\frac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\left(2\right)\)

Từ (1) và (2) => đpcm

 \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Ta có : \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)\(\Rightarrow\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\left(\frac{a+b}{c+d}\right)^3\)(1)

Ta lại có : \(\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^3=\left(\frac{b}{d}\right)^3=\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}\)(2)

Từ (1) và (2) \(\Rightarrowđpcm\)