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

a) \(\frac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\frac{\left(kb\right)^{2004}-b^{2004}}{\left(kb\right)^{2004}+b^{2004}}=\frac{k^{2004}b^{2004}-b^{2004}}{k^{2004}b^{2004}+b^{2004}}=\frac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\frac{k^{2004}-1}{k^{2004}+1}\)(1)

\(\frac{c^{2004}-d^{2004}}{d^{2004}+d^{2004}}=\frac{\left(kd\right)^{2004}-d^{2004}}{\left(kd\right)^{2004}+d^{2004}}=\frac{k^{2004}d^{2004}-d^{2004}}{k^{2004}d^{2004}+d^{2004}}=\frac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\frac{k^{2004}-1}{k^{2004}+1}\)(2)

Từ (1) và (2) => đpcm

b) \(\frac{a^{2005}}{b^{2005}}=\frac{\left(kb\right)^{2005}}{b^{2005}}=\frac{k^{2005}b^{2005}}{b^{2005}}=k^{2005}\)(1)

\(\frac{\left(a-c\right)^{2005}}{\left(b-d\right)^{2005}}=\frac{\left(kb-kd\right)^{2005}}{\left(b-d\right)^{2005}}=\frac{\left[k\left(b-d\right)\right]^{2005}}{\left(b-d\right)^{2005}}=\frac{k^{2005}\left(b-d\right)^{2005}}{\left(b-d\right)^{2005}}=k^{2005}\)(2)

Từ (1) và (2) => đpcm

Bạn tham khảo :

Ta có :

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

\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=1\)

\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2=0\)

\(\Rightarrow abc\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2\right)=abc.0\)

\(\Rightarrow a^2b+b^2c+a^2c+b^2a+c^2a+c^2b+2abc=0\)

\(\Rightarrow\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(a^2c+abc\right)+\left(c^2a+c^2b\right)=0\)

\(\Rightarrow ab\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Rightarrow\left(ab+bc+ac+c^2\right)\left(a+b\right)=0\)

\(\Rightarrow\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]\left(a+b\right)=0\)

\(\Rightarrow\left[b\left(a+c\right)+c\left(a+c\right)\right]\left(a+b\right)=0\)

\(\Rightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)

TH1 : \(a+c=0\)

\(\Rightarrow a=-c\)

\(\Rightarrow c^{2006}=a^{2006}\)

\(\Rightarrow P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)

\(=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)0\)

\(=0\)

CMTT đều có \(P=0\)

Vậy ...

hay quá cảm ơn nha nhưng có cách nào gọn hơn ko

Áp dụng tính chất của dãy tỉ số bằng nhau :

\(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=\frac{a-b}{2003-2004}=\frac{b-c}{2004-2005}=\frac{c-a}{2005-2003}\)

\(\Leftrightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)

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

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

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

Vậy ...

Đặt: \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=b\Rightarrow\hept{\begin{cases}a=2003b\\b=2004b\\c=2005b\end{cases}}\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(2003b-2004b\right)\left(2004b-2005b\right)=4.-b.-b=4b^2\)

\(\Rightarrow\left(c-a\right)^2=\left(2005b-2003b\right)^2=2k^2=4k^2\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(đpcm\right)\)

Đặt a/2003=b/2004=c/2005=k

Suy ra a=2003k, b=2004k, c=2005k            (*)

Thay (*) vào 4(a-b)(b-c) ta được:

4(a-b)(b-c)=4(2003k-2004k) (2004k-2005k)

              =4k(2003-2004).k(2004-2005)=4k² .-1.-1

              =4.k²                                                           (1)

Thay (*) vào (c-a)² ta được:

(c-a)² =(2005k-2003k)²

= k² (2005-2003)²

=k² .4                                                              (2)

Từ (1) và (2)

Suy ra ĐPCM

nha

Mình cũng học lớp 7 nhưng lần đầu mình thấy những loại toán này

coi \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\Rightarrow a=2003k;b=2004k;c=2005k\)

thay mấy cái trên vào 4(a-b)(b-c)và (c-a)²

Đặt \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\)

\(\Rightarrow a=2003k;b=2004k;c=2005k\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(2003k-2004k\right).\left(2004k-2005k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)(1)

     \(\left(c-a\right)^2=\left(2006k-2004k\right)^2=\left(2k\right)^2=4k^2\)(2)

Từ (1) và (2) 

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)

                                      đpcm

Tham khảo nhé~