\(a^3+6=-3a-2a^2\)

\(\Leftrightarrow a^3+2a^2+6+3a=0\)

\(\Leftrightarrow a^2\left(a+2\right)+3\left(a+2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(a^2+3\right)=0\)

\(\Leftrightarrow a+2=0\left(do.a^2+3>0\right)\)

<=>a=-2

thay a=-2 vào biểu thức ta được \(A=\frac{-2-1}{-2+3}=\frac{-3}{1}=-3\)

Ta có : a³+6=-3a-2a²

      <=> a³+6+3a+2a²=0

      <=>(a³+2a²)+(3a+6)=0

      <=>a²(a+2)+3(a+2)=0

      <=>(a²+3)(a+2)=0

      \(\hept{\begin{cases}a^2+3=0\\a+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a^2=-3\\a=-2\end{cases}\Leftrightarrow}\hept{\begin{cases}a\in\varnothing\\a=-2\end{cases}}}\)

Thay a=-2 vào biểu thức :

=> A= \(\frac{-2-2}{-2+3}=\frac{-4}{1}=-4\)

     \(a^3+6=-3a-2a^2\)

\(\Leftrightarrow\)\(a^3+2a^2+3a+6=0\)

\(\Leftrightarrow\)\(a^2\left(a+2\right)+3\left(a+2\right)=0\)

\(\Leftrightarrow\)\(\left(a+2\right)\left(a^2+3\right)=0\)

\(\Leftrightarrow\)\(a+2=0\)   (vì a² + 3 # 0)

\(\Leftrightarrow\)\(a=-2\)

Vậy   \(A=\frac{a-1}{a+3}=\frac{-2-1}{-2+3}=3\)

a³+6= -3a-2a².

->a=-2

\(\Leftrightarrow A=\frac{-2-1}{-2+3}=\frac{-3}{1}=-3\)

vậy A=-3

ĐKXD: a+3 khác 0 => a khác -3

Ta có a^3+6+3a+2a^2=0
         <=> a^2(a+2) + 3(a+2)=0
         <=> (a+2)(a^2+3)=0
=> a+2=0 <=> a= -2
Suy ra 
a-1/a+3= -2-1/-2+3=-3/1=-3 
 

a³+6= -3a-2a²

=>a³+2a²=-6-3a

=>a²(a+2)=-3.(2+a)

=>a²=-3(vô lý vì a²\(\ge\)0)

=>a²(a+2)=-3.(2+a)=0

=>2+a=0

=>a=0-2=-2

=>A=-2-1/-2+3=-3/1=-3

\(a^3+6=-3a-2a^2\)

\(\Leftrightarrow a^3+6+3a+2a^2=0\)

\(\Leftrightarrow a^2.\left(a+2\right)+3.\left(a+2\right)=0\)

\(\Leftrightarrow\left(a^2+3\right).\left(a+2\right)=0\Leftrightarrow a+2=0\Leftrightarrow a=-2\left(\text{vì }a^2+3\ge3\right)\)

Thay a=-2, vào A ta có:

\(A=\frac{-2-1}{-2+3}=-3\)

2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)