Đáp án:

$\begin{array}{l}
a)Dkxd:x\# 0;x\#  - 1;x\# 1\\
C = \left[ {\dfrac{2}{{3x}} - \dfrac{2}{{x + 1}}.\left( {\dfrac{{x + 1}}{{3x}} - x - 1} \right)} \right]:\dfrac{{x - 1}}{x}\\
 = \left( {\dfrac{2}{{3x}} - \dfrac{2}{{x + 1}}.\dfrac{{x + 1 - \left( {x + 1} \right)3x}}{{3x}}} \right).\dfrac{x}{{x - 1}}\\
 = \dfrac{{2.\left( {x + 1} \right) - 2.\left( {x + 1 - 3{x^2} - 3x} \right)}}{{3x.\left( {x + 1} \right)}}.\dfrac{x}{{x - 1}}\\
 = \dfrac{{2x + 2 - 2.\left( { - 3{x^2} - 2x + 1} \right)}}{{3.\left( {x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{2x + 2 + 6{x^2} + 4x - 2}}{{3\left( {x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{6{x^2} + 6x}}{{3\left( {x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{6x\left( {x + 1} \right)}}{{3\left( {x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{2x}}{{x - 1}}\\
b){x^2} + 3x + 2 = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow x =  - 2\left( {do:x\#  - 1} \right)\\
C = \dfrac{{2x}}{{x - 1}} = \dfrac{{2.\left( { - 2} \right)}}{{ - 2 - 3}} = \dfrac{4}{5}\\
c)C =  - \dfrac{2}{3}\\
 \Leftrightarrow \dfrac{{2x}}{{x - 1}} = \dfrac{{ - 2}}{3}\\
 \Leftrightarrow 3x =  - x + 1\\
 \Leftrightarrow 4x = 1\\
 \Leftrightarrow x = \dfrac{1}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{4}\\
d)C < 2\\
 \Leftrightarrow \dfrac{{2x}}{{x - 1}} < 2\\
 \Leftrightarrow \dfrac{{2x - 2\left( {x - 1} \right)}}{{x - 1}} < 0\\
 \Leftrightarrow \dfrac{2}{{x - 1}} < 0\\
 \Leftrightarrow x - 1 < 0\\
 \Leftrightarrow x < 1\\
Vậy\,x < 1;x\#  - 1;x\# 0
\end{array}$

a/ ĐKXĐ: $x\ne 0;\pm 1$

$C=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x-1}{x}\\=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-(x+1)\right)\right].\dfrac{x}{x-1}\\=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}.\dfrac{x+1}{3x}+\dfrac{2}{x+1}.(x+1)\right].\dfrac{x}{x-1}\\=\left[\dfrac{2}{3x}-\dfrac{2}{3x}+2\right].\dfrac{x}{x-1}\\=\dfrac{2x}{x-1}$

b/ $x^2+3x+2=0\\↔x^2+2x+x+2=0\\↔(x^2+2x)+(x+2)=0\\↔x(x+2)+(x+2)=0\\↔(x+1)(x+2)=0\\↔\left[\begin{array}{1}x+1=0\\x+2=0\end{array}\right.\\↔\left[\begin{array}{1}x=-1\\x=-2\end{array}\right.$

mà theo ĐKXĐ: $x\ne 0;\pm 1$

$→x=-2$ 

Với $x=-2$ (thỏa mãn điều kiện)

$→C=\dfrac{2.(-2)}{-2-1}=\dfrac{4}{3}$

c/ $C=\dfrac{-2}{3}\\→\dfrac{2x}{x-1}=\dfrac{-2}{3}\\↔\dfrac{6x}{3(x-1)}=\dfrac{-2(x-1)}{3(x-1)}\\↔6x=-2x+2\\↔8x=2\\↔x=\dfrac{1}{4}$

d/ $C<2\\→\dfrac{2x}{x-1}<2\\↔\dfrac{2x}{x-1}-2<0\\↔\dfrac{2x-2x+2}{x-1}<0\\↔\dfrac{2}{x-1}<0\\→x-1<0\\↔x<1$

Vậy $x<1$ thì $C<2$