1. Để A có nghĩa thì \(x^3-3x-2\ne0\)

\(\Rightarrow\left(x^3-x\right)-\left(2x-2\right)\ne0\)

\(\Rightarrow x\left(x^2-1\right)-2\left(x-1\right)\ne0\)

\(\Rightarrow x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\ne0\)

\(\Rightarrow\left(x^2+x-2\right)\left(x-1\right)\ne0\)

\(\Rightarrow\left(x^2-1+x-1\right)\left(x-1\right)\ne0\)

\(\Rightarrow\left[\left(x+1\right)\left(x-1\right)+\left(x-1\right)\right]\left(x-1\right)\ne0\)

\(\Rightarrow\left(x-1\right)^2\left(x+2\right)\ne0\)

\(\Rightarrow x\ne1;x\ne-2\)

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

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

3/ Để A < 1 \(\Leftrightarrow\frac{\left(x+1\right)^2}{x+2}< 1\Leftrightarrow\left(x+1\right)^2< x+2\)

                                                        \(\Leftrightarrow x^2+2x+1< x+2\)

                                                         \(\Leftrightarrow x^2+x< 1\)

                                                           \(\Leftrightarrow x.\left(x+1\right)< 1\)

Vậy .....

1. A có nghĩa khi \(x^3-3x-2\ne0\)

\(\Leftrightarrow x^3+x^2-x^2-x-2x-2\ne0\)

\(\Leftrightarrow x^2\left(x+1\right)-x\left(x+1\right)-2\left(x+1\right)\ne0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x-2\right)\ne0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x-2x-2\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow x-2\ne0\)(do \(\left(x+1\right)^2\ge0\)) \(\Leftrightarrow x\ne2\)

2. Ta có :

Tử = \(x^4-2x^2+1=x^4-x^3+x^3-x^2-x^2+x-x+1\)

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

=\(\left(x-1\right)\left(x^3+x^2-x-1\right)=\left(x-1\right)\left[x^2\left(x+1\right)-x\left(x+1\right)\right]\)

=\(\left(x-1\right)\left(x+1\right)\left(x^2-1\right)=\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)\)

\(=\left(x+1\right)^2\left(x-1\right)^2\)

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

3. \(A< 1\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}< 1\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Leftrightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)

\(\Leftrightarrow\frac{x^2-3x+3}{x-2}< 0\)ta có \(x^2-3x+3=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{3}{4}=\left(x-\frac{3}{4}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\)(1) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\)(Thỏa mãn)

Vậy x<2 thì A<1

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

A có nghĩa \(\Leftrightarrow x^3-3x-2\ne0\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

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

A<1\(\Rightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Rightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)

\(\Rightarrow\frac{x^2-3x+3}{x-2}< 0\Rightarrow x-2< 0\)vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy x<2 thỏa mãn yêu cầu A<1

Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy

Bài 4:

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

a) DK x khác +-1

b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)

c) x+1  phải thuộc Ước của 2=> x=(-3,-2,0))

1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)

                                      \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)

                                       \(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)

   Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa

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

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

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

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

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

 \(=\frac{x-2}{x+2}\)       

c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)             

\(\Leftrightarrow x-2=\left(x+2\right).0\)          

\(\Leftrightarrow x-2=0\)   

\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )

=> ko có gía trị nào của x để A=0

Bài 1: 

a) \(x+2\ne0\Leftrightarrow x\ne-2\)

\(x^2-4\ne0\Leftrightarrow x\ne+_-2\)

b) \(A=\frac{x}{x+2}+\frac{4-2x}{x^2-4}=\frac{x-2}{x+2}\)

c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Mà đk: x khác 2 

Vậy ko tồn tại giá trị nào của x để A=0

a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)

+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)

\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)

+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)

+) \(x+1\ne0\Leftrightarrow x\ne-1\)

+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)

\(\Leftrightarrow x\ne0;x\ne-2\)

+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)

Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)

a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)

\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)

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

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

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

\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)

b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)

Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Vậy ............................

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

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

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

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

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

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

A nguyên\(\Leftrightarrow3x⋮x-2\)

\(\Leftrightarrow3\left(x-2\right)+6⋮x-2\)

Mà \(\left(x-2\right)⋮x-2\Rightarrow6⋮x-2\)

\(\Rightarrow x-2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Lập bảng:

Vậy\(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

giúp mình mọi người ơi

a. \(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

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

c. Để \(A=-2\) thì \(\dfrac{3}{x-1}=-2=\dfrac{3}{\dfrac{-3}{2}}\\ \Leftrightarrow x-1=\dfrac{-3}{2}\\ \Leftrightarrow x=\dfrac{-1}{2}\left(\text{t/m ĐKXĐ}\right)\)

Vậy \(x=\dfrac{-1}{2}\) để phân thức nhận giá trị là -2.

a) Có: \(x^2-1=\left(x-1\right)\left(x+1\right)\)

ĐKXĐ là x ≠ 1; x ≠ -1

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

c) Theo đề ta có: \(\dfrac{3}{x-1}=2\)

\(\Rightarrow x-1=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{5}{2}\) (T/m ĐK)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(\dfrac{3x+3}{x^2-1}\)

\(=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{3}{x-1}\)

c) Để phân thức có giá trị bằng -2 thì \(\dfrac{3}{x-1}=-2\)

\(\Leftrightarrow x-1=\dfrac{-3}{2}\)

hay \(x=\dfrac{-1}{2}\)(nhận)

Vậy: Để phân thức có giá trị bằng -2 thì \(x=\dfrac{-1}{2}\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)

Ta có : \(P=\dfrac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{4}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x\left(x-3\right)+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x-x+6}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-3}{x-3}\)

b, Ta có : \(P=\dfrac{2x-3}{x-3}=\dfrac{2x-6+3}{x-3}=2+\dfrac{3}{x-3}\)

- Để P là số nguyên \(\Leftrightarrow x-3\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow x\in\left\{4;3;6;0\right\}\)

Vậy ...

a ĐKXĐ : \(x\ne2,x\ne3\)

\(\Rightarrow P=\dfrac{2x\left(x-3\right)+4-\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{x^2-5x+6}\)b Ta có P = \(\dfrac{2x^2-7x+6}{x^2-5x+6}=\dfrac{x^2-5x+6+x^2-2x}{x^2-5x+6}=1+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=1+\dfrac{x}{x-3}\)

Để P\(\in Z\) \(\Leftrightarrow1+\dfrac{x}{x-3}\in Z\) \(\Rightarrow\dfrac{x}{x-3}\in Z\) \(\Rightarrow x⋮x-3\) \(\Rightarrow x-3+3⋮x-3\)

\(\Rightarrow3⋮x-3\) \(\Rightarrow\left(x-3\right)\in\left\{-3;-1;1;3\right\}\) \(\Rightarrow x\in\left\{0;2;4;6\right\}\) 

Thử lại ta thấy đúng 

Vậy...