Vì a + b + c = 0

<=> (a + b + c)² = 0

<=> a² + b² + c² = -2(ab + bc + ca)

Khi đó \(\frac{9\left(a^2+b^2+c^2\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}=\frac{-18\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}\)

\(=\frac{-18\left(ab+bc+ca\right)}{-6\left(ab+bc+ca\right)}=3\)

a² - 6b² = ab

<=> (a + 2b)(a - 3b) = 0

<=> \(\orbr{\begin{cases}a=-2b\left(\text{loại}\right)\\a=3b\left(tm\right)\end{cases}}\)

Khi đó \(A=\frac{2ab}{a^2-7b^2}=\frac{6b^2}{2b^2}=3\)

\(\left(2a-b\right)^2=\left(5b\right)^2\)

\(\orbr{\begin{cases}2a=b+5b\\2a=b-5b\left(loai\right)\end{cases}}\)

a=3b

\(A=\frac{6b^2}{2b^2}=3\)

Ta có: \(a^2-6b^2=ab\)

=>\(a^2-ab-6b^2=0\)

=>\(a^2-3ab+2ab-6b^2=0\)

=>a(a-3b)+2b(a-3b)=0

=>(a-3b)(a+2b)=0

=>a=3b hoặc a=-2b

TH1: a=3b

\(A=\frac{2ab}{a^2-7b^2}\)

\(=\frac{2\cdot3b\cdot b}{\left(3b\right)^2-7b^2}=\frac{6b^2}{9b^2-7b^2}\)

\(=\frac{6b^2}{2b^2}=\frac62=3\)

TH2: a=-2b

\(A=\frac{2ab}{a^2-7b^2}\)

\(=\frac{2\cdot\left(-2b\right)\cdot b}{\left(-2b\right)^2-7b^2}=\frac{-4b^2}{4b^2-7b^2}\)

\(=\frac{-4b^2}{-3b^2}=\frac43\)

a^2-6b^2=-ab 

a^2+ab-6b^2=0 

a^2+3ab-2ab-6b^2=0

a(a+3b)-2b(a+3b)=0

(a+3b)(a-2b)=0 

suy ra a+3b=0 hoặc a-2b=0 

ta có a>b>0 nên a+3b=0 sẽ ko xảy ra 

suy ra a-2b=0 ,a=2b

thế vào đa thức M ta có M=2.2b.b/2.(2b)^2-3b^2 

M=4b^2/5b^2=4/5