Đáp án:

Giải thích các bước giải:

`A = [1 – ( x + 2)/(x+3)] : [(x – 2) / (x+3) + (x+3)/(2 - x) + (11x+8)/(x^2 +x - 6)] `

`ĐKXĐ:` $\begin{cases}x+3≠0\\2-x≠0\\x^2+x-6≠0\end{cases}$

`<=>`$\begin{cases}x+3≠0\\2-x≠0\\(x+3)(x-2)≠0\end{cases}$

`<=>`$\begin{cases}x≠-3\\x≠2\end{cases}$

`=>xne-3;2`

`B =[(2 + 3x)/(2 – 3x) - (36x^2) /(9x^2 - 4) - (2 – 3x)/(2+3x)]:[ (x^2 – x)/(2x^2 – 3x^3) ]`

`ĐKXĐ:9x^2-4=(3x-2)(3x-2)`

`2x^2-3x^3=x^2(2-3x)`

`x^2-x=x(x-1)`

$\begin{cases}2-3x≠0\\2+3x≠0\\x^2(2-3x)≠0\\x(x-1)≠0\end{cases}$

`<=>`$\begin{cases}x≠\dfrac{2}{3}\\x≠-\dfrac{2}{3}\\x≠0\\x≠1\end{cases}$

`=>xne+-2/3;0;1`

`C =[(1/(x + 1) - (x^3 -x) /(x^2 +1)] . [ 1/(x^2 + 2x + 1) – 1/ (x^2 – 1)] `

`ĐKXĐ:x^2+1ne0` (luôn đúng)

`x^2+2x+1=(x+1)^2`

`x^2-1=(x-1)(x+1)`

$\begin{cases}x+1≠0\\(x+1)^2≠0\\(x-1)(x+1)≠0\end{cases}$

`<=>`$\begin{cases}x≠-1\\x≠-1\end{cases}$

`=>xne+-1`

`D =[(x - 1)/(x-3) + (x -3) /(x + 3) - (4x -2)/(9 – x^2)]:[ 1 – (x + 1) /(3 + x)] ` $\begin{cases}x+3≠0\\x-3≠0\end{cases}$

`<=>`$\begin{cases}x≠-3\\x≠3\end{cases}$

`=>xne+-3`

`E =[(9 – 3x)/(x^2 + 4x - 5) - (x + 5) /(1 - x) - (x + 1)/(x + 5)]:[ (7x – 14) /(x^3 – 1)] `

`x^2+4x-5=(x-1)(x+5)`

`x^3-1=(x-1)(x^2+x+1)`

`x^2+x+1ne0` (luôn đúng)

`ĐKXĐ:` $\begin{cases}x-1≠0\\x+5≠0\\7x-14≠0\end{cases}$

`<=>`$\begin{cases}x≠1\\x≠-5\\x≠2\end{cases}$

`=>xne1;2;-5`