a: ĐKXĐ: x^3-3x-2<>0

=>x^3-x-2x-2<>0

=>x(x-1)(x+1)-2(x+1)<>0

=>(x+1)(x-2)(x+1)<>0

=>x<>2 và x<>-1

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

c: 

A<1

=>A-1<0

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

=>x-2<0

=>x<2

a) A đc xác định <=>2x+4\(\left\{{}\begin{matrix}2x+4\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne2\end{matrix}\right.\)

câu b bn quy đòng mẫu là đc

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

b) Ta có: \(A=\dfrac{x}{2x+4}+\dfrac{3x+2}{x^2-4}\)

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

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

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

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

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

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

c) Để A=0 thì \(\dfrac{x+2}{2\left(x-2\right)}=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2(Không thỏa mãn ĐKXĐ)

Vậy: Không có giá trị nào của x để A=0

a) ĐK: \(x\ne4,x\ne2;x\ne-2\)

b) \(A=\dfrac{x^3}{x-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

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

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

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

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

\(A=\dfrac{\left(x-1\right)\left(x^2-4\right)}{x^2-4}\)

\(A=x-1\)

c) \(A=0\) khi:

\(x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

d) A dương khi: \(A>0\)

\(x-1>0\)

\(\Leftrightarrow x>1\)

Kết hợp với đk: 

\(x>1,x\ne4,x\ne2\)

a) Phân thức A được xác định khi: \(x^2-1\ne0\Rightarrow\left(x-1\right)\left(x+1\right)\ne0\Rightarrow\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\)

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

Vây ĐKXĐ của A là \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

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

Vậy \(A=\dfrac{x+1}{x-1}\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

c) Ta có A=2 <-> \(\dfrac{x+1}{x-1}=2\Leftrightarrow x+1=2\left(x-1\right)\Leftrightarrow x+1=2x-2\)

\(\Leftrightarrow x+1-2x+2=0\Leftrightarrow3-x=0\Rightarrow x=3\)

Vậy khi x=3 thì A=2

a) Biểu thức A xác định `<=>x^2-1 ne 0 <=> (x-1)(x+1) ne 0 <=> x ne +-1`

b) `A=(x^2-3x-4)/(x^2 -1) = (x^2+x-4x-4)/(x^2-1) = (x(x+1)-4(x+1))/(x^2-1)`

`= ((x+1)(x-4))/((x+1)(x-1))=(x-4)/(x-1)`

c) `A` là số nguyên `<=> (x-4) vdots\ (x-1)`

`<=>[(x-1)-3] vdots\ (x-1)`

`<=> -3\ vdots\ (x-1)`

`<=> (x-1)\ in\ Ư(-3)`

`<=>(x-1)\ in\ {-3;-1;3;1}`

`<=>x\ in\ {-2;0;4;2}`

Vậy...

a: ĐKXĐ: x<>1; x<>-1

b: \(A=\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-4}{x-1}\)

c: Để A là số nguyên thì x-1-3 chia hết cho x-1

=>\(x-1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;0;4;-2\right\}\)

\(a,ĐK:x^2-1\ne0\)

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

\(\Rightarrow\left[{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

       Vậy ĐKXĐ của \(x\) là \(x\ne\pm1\)

\(b,\dfrac{x^2-3x-4}{x^2-1}=\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-4}{x-1}\)

\(c,\) Ta có: \(\dfrac{x-4}{x-1}=\dfrac{x-1-3}{x-1}=\dfrac{x-1}{x-1}-\dfrac{3}{x-1}=1-\dfrac{3}{x-1}\)

Để \(A\in Z\) thì \(\dfrac{3}{x-1}\in Z\)

\(\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Ta có bảng sau:

Vậy \(x\in\left\{2;0;4;-2\right\}\)

a: ĐKXĐ: x<>4; x<>-4

b: \(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x-1}{x+4}\)

c: Để A nguyên thì x+4-5 chia hết cho x+4

=>\(x+4\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-3;-5;1;-9\right\}\)

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

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

c: Thay x=2 vào A, ta được:

\(A=\dfrac{2+1}{2-1}=3\)

d: Để A=2 thì x+1=2x-2

=>-x=-3

hay x=3(nhận)

Lời giải:
a.

ĐKXĐ: $x\neq \pm 2$

b.
\(P=\left[\frac{4(x-2)}{(x+2)(x-2)}+\frac{3(x+2)}{(x+2)(x-2)}-\frac{5x+2}{(x-2)(x+2)}\right].\frac{x+2}{2}\)

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