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

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

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

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

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

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

a) \(\frac{2a^2-3a-2}{a^2-4}=2\)

\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)

\(\Rightarrow2a^2-3a-2=2a^2-4\)

\(\Rightarrow-3a-2=-4\)

\(\Rightarrow-3a=-2\Rightarrow a=\frac{2}{3}\)

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

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

\(\Rightarrow\frac{6a^2-6}{3a^2+10a+3}=2\)

\(\Rightarrow6a^2-6=2\left(3a^2+10a+3\right)\)

\(\Rightarrow6a^2-6=6a^2+20a+6\)

\(\Rightarrow-6=20a+6\Rightarrow20a=-12\)

\(\Rightarrow a=\frac{-3}{5}\)

a, \(\frac{2a^2-3a-2}{a^2-4}=2\)

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

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

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

\(\Leftrightarrow\frac{2a+1}{a+2}=2\)

\(\Leftrightarrow2a+1=2\left(a+2\right)\Leftrightarrow2a+1=2a+4\Leftrightarrow2a+1-2a-4=0\)

\(\Leftrightarrow-3\ne0\)(voli)

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

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

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

\(\Leftrightarrow6a^2-6=6a^2+20a+6\)

\(\Leftrightarrow6a^2-6-6a^2-20a-6=0\)

\(\Leftrightarrow-12-20a=0\)

\(\Leftrightarrow20a=-12\)

\(\Leftrightarrow a=-\frac{3}{5}\)

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, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

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.

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