a) \(ĐKXĐ:\hept{\begin{cases}a\ne1\\a\ne0\end{cases}}\)

\(M=\left(\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right)\div\frac{a^3+4a}{4a^2}\)

\(\Leftrightarrow M=\left(\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right):\frac{a^2+4}{4a}\)

\(\Leftrightarrow M=\frac{\left(a-1\right)^3-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)

\(\Leftrightarrow M=\frac{a^3-3a^2+3a-1-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)

\(\Leftrightarrow M=\frac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a^2}{a^2+4}\)

\(\Leftrightarrow M=\frac{4a^2}{a^2+4}\)

b) Ta có : \(\frac{4a^2}{a^2+4}=\frac{4\left(a^2+4\right)-16}{a^2+4}\)

\(=4-\frac{16}{a^2+4}\)

Để M đạt giá trị lớn nhất 

\(\Leftrightarrow\frac{16}{a^2+4}\)min

\(\Leftrightarrow a^2+4\)max

\(\Leftrightarrow a\)max

Vậy để M đạt giá trị lớn nhất thì a phải đạ giá trị lớn nhất.

\(x^3+x^2-2x=x^3+2x^2-x^2-2x=x^2\left(x+2\right)-x\left(x+2\right)=\left(x+2\right)x\left(x+1\right)\)

\(\text{nên phép chia:}x^3+x^2-2x\text{ cho:}x+2\text{ ko dư và có thương là:}x^2+x\)

Có gì sao sót thì thôi nha .

Ta có : 3.( x -3)²  + 2x = x . ( x + 1 ) + 9

<=> 3.x² -16.x + 27       = x. ( x + 1 ) + 9

<=> 3.x² - 16.x + 27       = x² + x+ 9

<=> 3.x² - 16.x +27 - x² -x - 9 = 0

<=>2.x² -17.x +18 = 0

<=> x=\(\frac{17\pm\sqrt{\left(-17\right)^2-144}}{4}\)

<=> x= \(\frac{17\pm\sqrt{145}}{4}\)

<=> x = \(\frac{17-\sqrt{145}}{4};\frac{17+\sqrt{145}}{4}\)

a, \(2+\frac{3}{x-5}=1\Leftrightarrow\frac{3}{x-5}=-1\)

\(\Leftrightarrow x-5=\frac{3}{-1}=-3\Leftrightarrow x=2\)

Vậy .............

b, ....................

\(\Leftrightarrow\frac{x-9}{x^2-3^2}-\frac{2}{x+3}=\frac{1}{x-3}\)

\(\Leftrightarrow\frac{x-9}{\left(x-3\right)\left(x+3\right)}-\frac{2x-6}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{x-9-2x+6-x+3}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{-2x}{\left(x-3\right)\left(x+3\right)}=0\Rightarrow-2x=0\Rightarrow x=0\)

Vậy .............