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:

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

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

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

\(a;b>0\Rightarrow3a+2b+1>1\)

\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\) đồng biến

Mà \(9a^2+b^2\ge2\sqrt{9a^2b^2}=6ab\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)\)

\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge2\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}log_{6ab+1}\left(3a+2b+1\right)=1\\3a=b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6ab+1=3a+2b+1\\b=3a\end{matrix}\right.\)

\(\Rightarrow18a^2+1=3a+6a+1\)

\(\Leftrightarrow18a^2-9a=0\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\)