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

  Ta có: \(\frac{a}{b}=\frac{c}{d}\)=> \(\frac{a}{c}=\frac{b}{d}=\frac{2005a}{2005c}=\frac{2006b}{2006d}=\frac{2006a}{2006c}=\frac{2007b}{2007d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

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

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

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

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

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

Mà \(3a>b>0\)nên \(3a-b>0\)

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

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

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

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