Đặt \(\frac{a}{b}=\frac{c}{d}=k\left(1\right)\Rightarrow a=bk;c=dk\)

Thay a và c vào tỉ số \(\frac{2005a-2006b}{2006c+2007d}=\frac{2005c-2006d}{2006a+2007b}\), ta có :

\(\frac{2005a-2006b}{2006c+2007d}=\frac{2005bk-2006b}{2006dk-2007d}=\frac{b\left(2005k-2006\right)}{d\left(2006k+2007\right)}\)

\(\frac{2005c-2006d}{2006a+2007b}=\frac{2005dk-2006d}{2006bk+2007b}=\frac{d\left(2005k-2006\right)}{b\left(2006k+2007\right)}\)

Mà \(\frac{b}{d}\ne\frac{d}{b}\left(b,d\in Z;b\ne d;b,d\ne0\right)\)

=> Sai đề

ko biết đúng ko nha, sai thì đừng chửi nhá

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a = b.k; c = d.k

\(\frac{2005a-2006b}{2006c+2007d}=\frac{2005b.k-2006b}{2006d.k+2007.d}=\frac{b\left(2005k-2006\right)}{d\left(2006k+2007\right)}=\frac{b}{d}.\frac{2005k-2006}{2006k+2007}\) (1)

\(\frac{2005c-2006d}{2006a+2007b}=\frac{2005d.k-2006d}{2006b.k+2007b}=\frac{d\left(2005k-2006\right)}{b\left(2006k+2007\right)}=\frac{d}{b}.\frac{2005k-2006}{2006k+2007}\) (2)

Từ (1)(2) => vế trái khác vế phải : Đề sai

Từ \(\dfrac{2005a-2006b}{2006c+2007d}=\dfrac{2005c-2006d}{2006a+2007b}\)

=> \(\dfrac{2005a-2006b}{2005c-2006d}=\dfrac{2006c+2007d}{2006a+2007b}\) (1)

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

=> \(\dfrac{2005a}{2005c}=\dfrac{2006b}{2006d}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{2005a}{2005c}=\dfrac{2006b}{2006d}=\dfrac{2005a+2006b}{2005c+2006d}\) (2)

Từ \(\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{2006a}{2006c}=\dfrac{2007d}{2007b}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{2006a}{2006c}=\dfrac{2007b}{2007d}=\dfrac{2006a-2007d}{2006c-2007b}\) (3)

Từ (1),(2),(3) => \(\dfrac{2005a-2006b}{2006c+2007d}=\dfrac{2005c-2006d}{2006a+2007b}\)

\(3a^2+2b^2=7ab\)

\(\Leftrightarrow3a^2+2b^2-7ab=0\)

\(\Leftrightarrow3a^2-6ab-ab+2b^2=0\)

\(\Leftrightarrow3a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(3a-b\right)\left(a-2b\right)=0\)

Mà \(3a>b>0\)nên \(3a-b>0\)

Vậy \(a-2b=0\Leftrightarrow a=2b\Leftrightarrow\frac{a}{2}=\frac{b}{1}\)

Đặt \(\frac{a}{2}=\frac{b}{1}=k\Rightarrow\hept{\begin{cases}a=2k\\b=k\end{cases}}\)

\(\Rightarrow P=\frac{2005.2k-2006.k}{2006.2k+2007.k}=\frac{2004k}{6019k}=\frac{2004}{6019}\)

Mk sẽ làm theo đề bài mà bạn nói dưới bình luận câu trả lời của bạn @Hồng Phúc Nguyễn.

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)

a) Thay (1) vào đề:

\(VT=\dfrac{a+2006b}{a-2006b}=\dfrac{bk+2006b}{bk-2006b}=\dfrac{b\left(k+2006\right)}{b\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)

\(VP=\dfrac{c+2006d}{c-2006d}=\dfrac{dk+2006d}{dk-2006d}=\dfrac{d\left(k+2006\right)}{d\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}.\)

b) Thay (1) vào đề:

\(VT=\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{2006\left(bk+dk\right)}{2006bk}=\dfrac{bk+dk}{bk}=\dfrac{k\left(b+d\right)}{bk}=\dfrac{b+d}{b}\)

\(VP=\dfrac{b+d}{b}\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{b+d}{b}\rightarrowđpcm\).

\(\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}\)

\(\Leftrightarrow\)(a+2006b)(c-2006d)=(c+2006d)(a-2006b)

a(c-2006d)+2006b(c-2006d)=c(a-2006b)+2006d(a-2006b)

ac-2006ad+2006bc-4024036bd=ac-2006bc+2006ad-4024036bd

(ac-2006ad+2006bc-402436bd)-(ac-2006bc+2006ad-4024036bd=0

Suy ra 2 đẳng thức trên =nhau

Đề là gì vậy bạn???????!!!!!!!!!!!!!!!!