a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-5x}{x^2-1}\right)\cdot\frac{x-3}{x}\left(x\ne\pm1;x\ne0\right)\)

\(\Leftrightarrow A=\left[\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-5x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{x-3}{x}\)

\(\Leftrightarrow A=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-5x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{x-3}{x}\)

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

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

Vậy \(A=\frac{x-3}{x+1}\left(x\ne\pm1;x\ne0\right)\)

b) \(A=\frac{x-3}{x+1}\left(x\ne\pm1;x\ne0\right)\)

Để A nhận giá trị nguyên thì x-3 chia hết chi x+1

=> (x+1)-4 chia hết chi x+1

=> 4 chia hết cho x+1

x nguyên => x+1 nguyên => x+1 thuộc Ư (4)={-4;-2;-1;1;2;4}
Ta có bảng

Vậy x={-5;-3;-2;3} thì A đạt giá trị nguyên

c) I3x-1I=5

\(\Rightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{-4}{3}\end{cases}}}\)

Đên đây thay vào rồi tính nhé

a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne0\end{cases}}\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-5x}{x^2-1}\right)\cdot\frac{x-3}{x}\)

\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-5x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)

\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1+x^2-5x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)

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

\(\Leftrightarrow A=\frac{x-3}{x+1}\)

b) Để \(A\inℤ\)

\(\Leftrightarrow x-3⋮x+1\)

\(\Leftrightarrow x+1-4⋮x+1\)

\(\Leftrightarrow4⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow x\in\left\{0;-2;-3;1;3;-5\right\}\)

Mà \(x\ne0;x\ne1\)

\(\Leftrightarrow x\in\left\{-2;-3;3;-5\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-3;3;-5\right\}\)

c) Khi \(\left|3x-1\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)

Vì khi x = 2 hoặc x = -4/3 thì x không thuộc tập hợp các giá trị làm cho A nguyên

Vậy khi |3x - 1| = 5 thì để cho A nguyên \(\Leftrightarrow x\in\varnothing\)

PLEASE HELP ME !!!

a) Đk \(x>0\)và \(x\ne4\)

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

=\(\frac{2\sqrt{x}}{x-4}\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

=\(\frac{2}{\sqrt{x}+2}\)

b) Để \(\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)

\(\Leftrightarrow\frac{4-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

\(\Leftrightarrow\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

Vì \(2\left(\sqrt{x}+2\right)>0\)

mà\(\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

nên \(-\sqrt{x}+2>0\)\(\Leftrightarrow x< 4\)

Vậy vs \(0< x< 4\)thì \(A>\frac{1}{2}\)

c) Để B=\(\frac{14}{3\left(\sqrt{x}+2\right)}\)đạt giá trị nguyên thì

\(3\left(\sqrt{x}+2\right)\inƯ\left(14\right)=\left(1;2;7\right)\)

+)\(3\left(\sqrt{x}+2\right)=1\)\(\Leftrightarrow\sqrt{x}+2=\frac{1}{3}\)\(\Rightarrow\)vô lý

+)\(3\left(\sqrt{x}+2\right)=2\)\(\Leftrightarrow\sqrt{x}+2=\frac{2}{3}\)\(\Rightarrow\)vô lý

+)\(3\left(\sqrt{x}+2\right)=7\)\(\Leftrightarrow\sqrt{x}+2=\frac{7}{3}\)\(\Leftrightarrow\sqrt{x}=\frac{1}{3}\)\(\Rightarrow x=\frac{1}{9}\)(TM đk)

Vậy vs \(x=\frac{1}{9}\)thì B có giá trị nguyên

ĐK để phân thức XĐ : x khác 1 và x> 0

 Đặt \(B=\left(\frac{\left(\sqrt{x}+2\right)\left(x-1\right)-\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+1\right)}{\left(x+2\sqrt{x}+1\right)\left(x-1\right)}\right)\) ( Đây là mình vừa đặt vừa làm mẫu thức chung nhe)

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

=>\(B=\frac{2\sqrt{x}+2x}{\left(\sqrt{x}+1\right)^2\left(x-1\right)}=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(x-1\right)}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)

A = \(B:\frac{\sqrt{x}}{\sqrt{x+1}}=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{2}{x-1}\)

B, Bạn tự làm ý B nhe

HD để A nguyên => x - 1 thuộc ước của 2 mà 2 có các ước là +-1 và +-2

(+) với x-1 = 2 => x = 3

............................

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

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

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

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

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

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

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

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

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

A = \(\frac{x+1}{x-1}\) (Đk: \(x-1\ge0\) => x \(\ge\)1)

b) Ta có: A = \(\frac{x+1}{x-1}=\frac{\left(x-1\right)+2}{x-1}=1+\frac{2}{x-1}\)

Để A \(\in\)Z <=> 2 \(⋮\)x - 1

<=> x - 1 \(\in\)Ư(2) = {1; -1; 2; -2}

<=> x \(\in\){2; 0; 3; -1}

c) Ta có: A < 0

=> \(\frac{x+1}{x-1}< 0\)

=> \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)

=> \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(loại) hoặc \(\hept{\begin{cases}x>-1\\x< 1\end{cases}}\) 

=> -1 < x < 1

Edogawa Conan

Thiếu dòng đầu  \(ĐKXĐ:\hept{\begin{cases}x\ne1\\x\ne-2\\x\ne0\end{cases}}\)

ĐKXĐ : \(\) x # +1 ; x # - 1 ; x # -2 ; x # 0 ; x # 2

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

  \(=\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}.\frac{x}{1-x}\)

  \(=\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{1-x}\)

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

  \(=\frac{3x^2+3x-3}{x^2+x-2}-\frac{2x^2-5}{x^2+x-2}\)

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

   \(\frac{x+1}{x-1}\)

b. Ta có:  \(A=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)

Để A nhận giá trị nguyên thì: \(2⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(2\right)\)

  +) x - 1 = 1 => x = 2   (loại)

  +) x - 1 = 2 => x = 3  

  +) x - 1 = -1 => x = 0  (loại)

  +) x - 1 = -2 => x = -1    (loại)

Vậy x = 3 là giá trị cần tìm.

c.  \(A< 0\Leftrightarrow\frac{x+1}{x-1}< 0\)

\(\Leftrightarrow\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)   hoặc \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>-1\\x< 1\end{cases}}\)    hoặc   \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(vô lý)

Vậy \(-1< x< 1\) và x # 0 là giá trị cần tìm

\(ĐKXĐ:x\ne\pm1\)

a) \(A=\left(\frac{1}{1-x}+\frac{2}{1+x}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)

\(=\left(\frac{\left(1+x\right)}{\left(1+x\right)\left(1-x\right)}+\frac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)

\(=\frac{1+x+2-2x-5+x}{1-x^2}:\frac{2x-1}{1-x^2}\)

\(=\frac{8}{1-x^2}.\frac{1-x^2}{2x-1}=\frac{8}{2x-1}\)

b) Để A nguyên thì \(\frac{8}{2x-1}\inℤ\)

\(\Leftrightarrow8⋮2x-1\Rightarrow2x-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Mà dễ thấy 2x - 1 lẻ nên\(2x-1\in\left\{\pm1\right\}\)

+) \(2x-1=1\Rightarrow x=1\left(ktmđkxđ\right)\)

+) \(2x-1=-1\Rightarrow x=0\left(tmđkxđ\right)\)

Vậy x nguyên bằng 0 thì A nguyên

c) \(\left|A\right|=A\Leftrightarrow A\ge0\)

\(\Rightarrow\frac{8}{2x-1}\ge0\Rightarrow2x-1>0\Leftrightarrow x>\frac{1}{2}\)

Vậy \(x>\frac{1}{2}\)thì |A| = A

a, \(A=\left(\frac{1}{1-x}+\frac{2}{1+x}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)

\(\Leftrightarrow A=\left(\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2-2x}{\left(1-x\right)\left(1+x\right)}-\frac{5-x}{\left(1-x\right)\left(1+x\right)}\right):\frac{\left(x+1\right)\left(x-1\right)}{2x-1}\)

\(\Leftrightarrow A=\frac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2x-1}\)

\(\Leftrightarrow A=\frac{-2\left(1-x^2\right)}{\left(1-x^2\right)\left(2x-1\right)}=\frac{2}{2x-1}\)

Vậy \(A=\frac{2}{2x-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)

b) \(A=\frac{2}{2x-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)

Để A nhận giá trị nguyên thì 2 chia hết cho 2x-1

Mà x nguyên => 2x-1 nguyên

=> 2x-1 thuộc Ư (2)={-2;-1;1;2}
Ta có bảng

Đối chiếu điều kiện

=> x=0