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.

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