Ta có \(\frac{3a-2b}{4}=\frac{2c-4a}{3}=\frac{4b-3c}{2}\)

=> \(\frac{12a-8b}{16}=\frac{6c-12a}{9}=\frac{8b-6c}{4}=\frac{12a-8b+6c-12a+8b-6c}{16+9+4}=\frac{0}{29}=0\)

=> \(\hept{\begin{cases}12a-8b=0\\6c-12a=0\\8b-6c=0\end{cases}}\Rightarrow\hept{\begin{cases}12a=8b\\6c=12a\\8b=6c\end{cases}}\Rightarrow\hept{\begin{cases}3a=2b\\2c=4a\\4b=3c\end{cases}}\Rightarrow\hept{\begin{cases}\frac{a}{2}=\frac{b}{3}\\\frac{a}{2}=\frac{c}{4}\\\frac{c}{4}=\frac{b}{3}\end{cases}}\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

=> \(\frac{\left|a\right|}{\left|2\right|}=\frac{\left|b\right|}{\left|3\right|}=\frac{\left|c\right|}{\left|4\right|}=\frac{\left|a\right|-\left|b\right|-\left|c\right|}{\left|2\right|-\left|3\right|-\left|4\right|}=\frac{-10}{2-3-4}=\frac{-10}{-5}=2\)

=> \(\hept{\begin{cases}a=\pm4\\b=\pm6\\c=\pm8\end{cases}}\)Vì mẫu số cùng dấu => Tử số cùng dấu

=> Các cặp (a;b;c) tìm được là (4;6;8) ; (-4;-6;-8)

\(A-B+C=2a^2-3ab+4b^2-3a^2-4ab+b^2+a^2+2ab+b^2\)

\(\Rightarrow A-B+C=-5ab+6b^2\)

A−B+C=−5ab+6b2

a, \(\left(a^2+b^2-2ab+2a-2b+1\right)+\left(b^2-2b+1\right)=0\)

=> \(\left(a-b+1\right)^2+\left(b-1\right)^2=0\)

Mà \(\left(a-b+1\right)^2\ge0,\left(b-1\right)^2\ge0\)

=> \(\hept{\begin{cases}a-b+1=0\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}}\)

b,Tương tự 

\(\left(a-2b+1\right)^2+\left(b-1\right)^2=0\)

=>\(\hept{\begin{cases}a=1\\b=1\end{cases}}\)

Bất đẳng thức đã cho tương đương:

\(\dfrac{6a^3b+4ab^3+a^2b^2+4b^4}{3a^4+14a^2b^2+8b^4}\le\dfrac{3}{5}\)

\(\Leftrightarrow\left(a-b\right)^2\left(3a-2b\right)^2\ge0\) (luôn đúng).

\(a=\dfrac{1}{10}\)

\(b=\dfrac{1}{20}\)