Điều kiện xác định của \(P\)là: 

\(\hept{\begin{cases}x^2+2x+1\ne0\\x^2-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\x\ne0\end{cases}}\)

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

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

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

Để \(P\)nguyên mà \(x\)nguyên suy ra \(x+1\inƯ\left(2\right)=\left\{-2,-1,1,2\right\}\Leftrightarrow x\in\left\{-3,-2,0,1\right\}\)

Đối chiếu điều kiện ta được \(x\in\left\{-3,-2\right\}\)thỏa mãn. 

a) Điều kiện: \(x\ne0;x\ne1\)

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

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

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

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

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

c) Thay: \(x=2\)vào \(\frac{1}{x+1}\)ta có: \(A=\frac{1}{2+1}=\frac{1}{3}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

b)

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

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

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

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

c) \(A=\frac{1}{x+1}=\frac{1}{2+1}=\frac{1}{3}\)

Vậy \(A=\frac{1}{3}\)

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

với ĐKXĐ ta có

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

=\(\frac{4x}{\left(x+1\right)\left(x-1\right)}\times\frac{7\left(x-1\right)}{2x}\)

=\(\frac{14}{x+1}\)

b, x=6(t/m)

khi x=6 thì A=\(\frac{14}{6+1}=2\)

c,A=7<=>\(\frac{14}{x+1}=7\)

         \(\Leftrightarrow7x+7=14\)

           \(\Leftrightarrow7x=7\Leftrightarrow x=1\left(loại\right)\)

Vậy ko có giá trị x để A=7

a) \(ĐKXĐ:x\ne1\)

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

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

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

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

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

\(=\frac{1}{x-1}\)

c) Với \(\forall x\)(\(x\ne1\)) thì biểu thức được xác định .

P/s : Theo mik câu c nên chuyển thành : Tìm x để biểu thức đạt giá trị nguyên.

Tại thấy câu c k khác j câu a !

ĐKXĐ:\(x\ne\pm2;x\ne0;x\ne3\)

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

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

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

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

\(=\frac{4x^2}{x-3}\)

b

Tại x=-2 thì biểu thức trên không xác định

Vậy A không xác định tại x=-2

c

\(A>0\Leftrightarrow\frac{4x^2}{x-3}>0\) mà \(4x^2>0\) ( nên nhớ là ĐKXĐ x khác 0 ) nên x-3 >0 hay x > 3

d

\(\left|x-7\right|=4\Leftrightarrow x-7=4\left(h\right)x-7=-4\)

\(\Leftrightarrow x=11\left(h\right)x=3\)

Loại trường hợp x=3 bạn thay x=11 vào tính tiếp nha !!!!!

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

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

a) ĐKXĐ:

\(\hept{\begin{cases}x^2+2x+1\ne0\\x^2-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2\ne0\\\left(x-1\right)\left(x+1\right)\ne0\\x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x\ne1;x\ne-1\\x\ne0\end{cases}}}\)

<=> x khác -1

       x khác 1; x khác -1

       x khác 0 

<=> x khác -1;1;0

Vậy ĐKXĐ là x khác -1;1;0

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

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

MTC: (x+1)^2(x-1)

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

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

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

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

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

\(\Rightarrow P=-\frac{x}{\left(x+1\right)^2}\) (tmđkxđ)

c)

\(P=-\frac{x}{\left(x+1\right)^2}=-\frac{x+1-1}{\left(x+1\right)\left(x+1\right)}=-\frac{x+1}{x+1}-\frac{1}{x+1}=-1-\frac{1}{x+1}\) ( ĐKXĐ là x khác -1;1;0) \(\left(P\in Z\right)\)

\(P\in Z\Leftrightarrow\frac{-1}{x+1}\)

Nên x+1 thuộc Ư(-1)={1;-1)

x+1=1=>x=1-1=0 ( o t/m đk)

x+1=-1=>x=-1-1=-2( (t/m đk) 

<=> x thuộc -2 thì gt của BT P là số nguyên 

bài 1 :

tự làm