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

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

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

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

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

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

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 
 

a: ĐKXĐ: a∉{-1/3;-3}

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

=>\(\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)

=>\(2\left(3a+1\right)\left(a+3\right)=\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)\)

=>\(2\left(3a^2+9a+a+3\right)=3a^2+9a-a-3+3a^2-9a+a-3\)

=>\(6a^2+20a+6=6a^2-6\)

=>20a=-12

=>a=-3/5(nhận)

b: ĐKXĐ: a∉{5/2;2/3}

\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)

=>\(\frac{2a-5-4}{2a-5}+\frac{3a-2+2}{3a-2}=2\)

=>\(1-\frac{4}{2a-5}+1+\frac{2}{3a-2}=2\)

=>\(\frac{2}{3a-2}=\frac{4}{2a-5}\)

=>\(\frac{4}{6a-4}=\frac{4}{2a-5}\)

=>6a-4=2a-5

=>4a=-1

=>a=-1/4(nhận)

c: ĐKXĐ: a<>-3

\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)

=>\(\frac{3a-1}{4a+12}+\frac{7a+2}{6a+18}=\frac{10}{3}-2=\frac43\)

=>\(\frac{3\left(3a-1\right)}{12\left(a+3\right)}+\frac{2\left(7a+2\right)}{12\left(a+3\right)}=\frac43\)

=>\(\frac{9a-3+14a+4}{12\left(a+3\right)}=\frac{4\cdot4\cdot\left(a+3\right)}{12\left(a+3\right)}\)

=>23a+1=16(a+3)=16a+48

=>7a=47

=>a=47/7(nhận)

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

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

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

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

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.

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

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

=>\(\left[\begin{array}{l}a^2-3a+1=1\\ a^2-3a+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2-3a=0\\ a^2-3a+2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}a\left(a-3\right)=0\\ \left(a-1\right)\left(a-2\right)=0\end{array}\right.\Rightarrow a\in\left\lbrace0;1;2;3\right\rbrace\)

ĐKXĐ: a<>2

=>a∈{0;1;3}

\(A=\frac{2a^3-12a^2+17a-2}{a-2}\)

\(=\frac{2a^3-4a^2-8a^2+16a+a-2}{a-2}=\frac{2a^2\left(a-2\right)-8a\left(a-2\right)+\left(a-2\right)}{a-2}\)

\(=2a^2-8a+1\)

Khi a=0 thì \(A=2a^2-8a+1=2\cdot0^2-8\cdot0+1=1\)

Khi a=1 thì \(A=2a^2-8a+1=2\cdot1^2-8\cdot1+1=2-8+1=3-8=-5\)

Khi a=3 thì \(A=2a^2-8a+1=2\cdot3^2-8\cdot3+1=18-24+1=19-24=-5\)