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![](https://rs.olm.vn/images/avt/0.png?1311)
1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)
1/(x+3)+1/(x-3)=
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(Q=\left(\frac{1}{x+1}+\frac{6x+3}{x^3+1}-\frac{2}{x^2-x+1}\right):\left(x+2\right)\)\(\left(ĐKXĐ:x\ne-1\right)\)
\(Q=\left(\frac{x^2-x+1}{x^3+1}+\frac{6x+3}{x^3+1}-\frac{2\left(x+1\right)}{x^3+1}\right):\left(x+2\right)\)
\(Q=\left(x^2-x+1+6x+3-2x-2\right):\left(x+2\right)\)
\(Q=\left(x^2+3x+2\right):\left(x+2\right)\)
\(Q=\left(x+1\right)\left(x+2\right):\left(x+2\right)\)
\(Q=x+1\)
b)Tại \(Q=\frac{1}{3}\)ta được:\(\frac{1}{3}=x+1\)
\(\Rightarrow x=-\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2=2x^3-\frac{3}{2}x^2+2\)
\(2x^2-10x-3x-2x^2=26\)
-13x=26
x=-2