Đặt:

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

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

\(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^{2014}=\left(\dfrac{bk+b}{dk+d}\right)^{2014}=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2014}=\left(\dfrac{b}{d}\right)^{2014}\)\(\Rightarrow\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{bk^{2014}+b^{2014}}{dk^{2014}+d^{2014}}=\dfrac{b\left(k^{2014}+b^{2013}\right)}{d\left(k^{2014}+d^{2013}\right)}\)

2 cái này thấy nó ko giống nhau lắm:v

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có:+) \(\left(\dfrac{a+b}{c+d}\right)^{2014}=\left(\dfrac{bk+b}{dk+d}\right)^{2014}=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2014}=\left(\dfrac{b}{d}\right)^{2014}\) (1)

+) \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}.k^{2014}+b^{2014}}{d^{2014}.k^{2014}+d^{2014}}\)

\(=\dfrac{b^{2014}.\left(k^{2014}+1\right)}{d^{2014}.\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\) (2)

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

Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\left(\dfrac{a+b}{c+d}\right)^{2014 }\)
=>\(\dfrac{a^{2014}}{c^{2014}}=\dfrac{b^{2014}}{d^{2014}}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\)
=>\(\left(\dfrac{a+b}{c+d}\right)^{2014}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\)(đpcm)

\(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\\ \dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{\left(bk\right)^{2014}+b^{2014}}{\left(dk\right)^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}\\ \left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{b^{2014}}{d^{2014}}\\ \RightarrowĐPCM\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

Xét \(VT=\left(\dfrac{a-b}{c-d}\right)^{2014}=\left(\dfrac{bk-b}{dk-d}\right)^{2014}=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}\left(1\right)\)

Xét \(VP=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\dfrac{b^{2014}k^{2014}+b^{2014}}{d^{2014}k^{2014}+d^{2014}}=\dfrac{b^{2014}\left(k^{2014}+1\right)}{d^{2014}\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) ta có ĐPCM

Đặt : \(\dfrac{a}{b}=\dfrac{c}{d}=k\) (k khác 0)

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

Khi đó:

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

\(=\left(\dfrac{b.\left(k-1\right)}{d.\left(k-1\right)}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}\) (1)

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

\(=\dfrac{b^{2014}.\left(k^{2014}+1\right)}{d^{2014}.\left(k^{2014}+1\right)}=\dfrac{b^{2014}}{d^{2014}}=\left(\dfrac{b}{d}\right)^{2014}\) (2)

Từ (1) và (2) suy ra

(đ.p.c.m)

Tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) có thể viết \(\dfrac{a}{c}=\dfrac{b}{d}\). Theo tính chất của dãy tỉ số bằng nhau ta có: \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\) hay nâng lên lũy thừa 2014:

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

Áp dụng lần nữa tính chất của tỉ số bằng nhau sẽ được:

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

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\) Theo t/c dãy tỷ số băng nhau ta có

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\left(\frac{a-b}{c-d}\right)^{2014}=\frac{a^{2014}}{c^{2014}}=\frac{b^{2014}}{d^{2014}}\) Theo t/c dãy tỷ số bằng nhau

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

v~ cay wa đề mình là (a+b)^2014/(c+d)^2014= a^2014+b^2014/ c^2014+d^2014