Đáp án + Giải thích các bước giải:

`a)` `A=(2/(x-1)+3/(x+1)-(2x+2)/(x^2-1))*(x^2+x)/(3x+9)(xne-3,xnepm1)`

`=([2(x+1)]/[(x-1)(x+1)]+[3(x-1)]/[(x-1)(x+1)]-(2x+2)/[(x-1)(x+1)])*[x(x+1)]/[3(x+3)]`

`= [2(x+1)+3(x-1)-2x-2]/[(x-1)(x+1)]*[x(x+1)]/[3(x+3)]`

`= (2x + 2 + 3x - 3 - 2x - 2)/[(x-1)(x+1)]*[x(x+1)]/[3(x+3)]`

`= (3x-3)/[(x-1)(x+1)]*[x(x+1)]/[3(x+3)]`

`= [3(x-1)]/[(x-1)(x+1)]*[x(x+1)]/[3(x+3)]=[3x(x-1)(x+1)]/[3(x-1)(x+1)(x+3)]`

`= x/(x+3)(đpcm)`

`b) |x-2|=1 `=> `\(\left[ \begin{array}{l}x-2=1\\x-2=-1\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\) 

Khi `x = 3(tmđk)` thì giá trị của biểu thức `A` là :

`A=x/(x+3)=3/(3+3)=3/6=1/2`

Khi `x=1(ktmđk)` thì không thể tìm được giá trị của biểu thức `A`

`c)` `x/(x+3)<1<=>x/(x+3)-1<0<=> x/(x+3)-(x+3)/(x+3)<0` $\\$ `<=>(x-x-3)/(x+3)<0 <=>-3/(x+3)<0<=>3/(x+3)>0<=>x+3>0<=>x> -3` $\\$

`=> x> -3,x ne pm 1,xne -3`

$\begin{array}{l} a)A = \left( {\dfrac{2}{{x - 1}} + \dfrac{3}{{x + 1}} - \dfrac{{2x + 2}}{{{x^2} - 1}}} \right).\dfrac{{{x^2} + x}}{{3x + 9}}\\ A = \dfrac{{2\left( {x + 1} \right) + 3\left( {x - 1} \right) - 2x - 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x\left( {x + 1} \right)}}{{3\left( {x + 3} \right)}}\\ A = \dfrac{{3x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x\left( {x + 1} \right)}}{{3\left( {x + 3} \right)}} = \dfrac{x}{{x + 3}}\\ b)\left| {x - 2} \right| = 1 \Leftrightarrow \left[ \begin{array}{l} x - 2 = 1\\ x - 2 =  - 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 1 \end{array} \right.\\  \Rightarrow \left[ \begin{array}{l} A = \dfrac{x}{{x + 3}} = \dfrac{3}{{3 + 3}} = \dfrac{1}{2}\\ A(L\,do\,x \ne 1) \end{array} \right.\\ c)A = \dfrac{x}{{x + 3}} < 1 \Leftrightarrow \dfrac{{x - x - 3}}{{x + 3}} < 0 \Leftrightarrow \dfrac{{ - 3}}{{x + 3}} < 0 \Leftrightarrow x + 3 > 0 \Leftrightarrow x >  - 3 \end{array}$

$\Rightarrow x> -3; x \ne \pm 1$