a) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)

\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)

\(=1+\left[\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)

\(=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)

\(=1+\frac{4x-2x^2}{x+1}.\frac{1}{x\left(x-2\right)}\)

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

b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:

\(|x-\frac{3}{4}|=\frac{5}{4}\)

*TH1: 

\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)

*TH2:

\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)

\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)

c,

\(Q=\frac{x-1}{x+1}=1-\frac{2}{x+1}\)

Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!

ĐKXĐ : \(x\ne\pm3\)

a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)

\(A=\left(\frac{-2x\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{\left(x+1\right)\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}+\frac{x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3}{x+3}-\frac{x-1}{x+3}\right)\)

\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)

\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)

\(A=\frac{-2\left(4x+1\right)\left(x+3\right)}{\left(3-x\right)\left(3+x\right)4}\)

\(A=\frac{-\left(4x+1\right)}{2\left(3-x\right)}\)

\(A=\frac{4x+1}{2\left(x-3\right)}\)

b) \(\left|x-5\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)

Mà ĐKXĐ x khác 3 => ta xét x = 7

\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)

c) Để A nguyên thì 4x + 1 ⋮ 2x - 3

<=> 4x - 6 + 7 ⋮ 2x - 3

<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3

Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3

=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }

=> x thuộc { 2; 1; 5; -2 }

Vậy .....

a)   ĐKXĐ: \(x\ne\pm3\)

   \(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)

 \(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)

\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)

\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)

b)

  Có 2 trường hợp:

T.Hợp 1:

               \(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)

thay vào A ta được: A=\(-\frac{13}{8}\)

T.Hợp 2:

          \(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)

Vậy không tồn tại giá trị của A tại x=3

Vậy với x=7 thì A=-13/8

c)

      \(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)

Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)

Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .

Vậy không có giá trị nguyên nào của x để A nguyên

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)

\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)

\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)

\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)

...................... 

tìm giá trị x nguyên để A nguyên đi

\(A=\left(\frac{x^2-16}{x-4}-1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)ĐK : \(x\ne3;-1;4\)

\(=\left(\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)

\(=\left(x-3\right):\left(\frac{x^2-x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)=\left(x-3\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x-1\right)}\right)\)thơm thế :))

\(=\left(x-3\right):\left(\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-1\right)}\right)=\left(x-3\right).\frac{x-1}{x+3}=\frac{\left(x-3\right)\left(x-1\right)}{x+3}\)

1) đk: \(x\ne\left\{-1;3;4\right\}\)

Ta có:

\(A=\left(\frac{x^2-16}{x-4}-1\right)\div\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)

\(A=\left[\frac{\left(x-4\right)\left(x+4\right)}{x-4}-1\right]\div\frac{\left(x-2\right)\left(x+1\right)+\left(x+3\right)\left(x-3\right)+x+2-x^2}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+4-1\right)\div\frac{x^2-x-2+x^2-9-x^2+x+2}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+3\right)\div\frac{x^2-9}{\left(x+1\right)\left(x-3\right)}\)

\(A=\left(x+3\right)\cdot\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=x+1\)

2) Ta có: \(\frac{A}{x^2+x+1}=\frac{x+1}{x^2+x+1}\)

Để \(\frac{A}{x^2+x+1}\) nguyên thì \(\left(x+1\right)⋮\left(x^2+x+1\right)\Leftrightarrow\left(x+1\right)^2⋮\left(x^2+x+1\right)\)

\(\Rightarrow\left(x+1\right)^2-\left(x^2+x+1\right)⋮\left(x^2+x+1\right)\)

\(\Rightarrow x⋮\left(x^2+x+1\right)\Rightarrow1⋮x^2+x+1\)

\(\Rightarrow x^2+x+1\in\left\{-1;1\right\}\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\orbr{\begin{cases}x=-1\left(ktm\right)\\x=0\left(tm\right)\end{cases}}\)

Vậy x = 0