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

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

\(P=\frac{x+2-x+2+x^2-4}{x^2-4}:\frac{1}{x^2-4}\)

\(P=\frac{x^2}{x^2-4}.\frac{x^2-4}{1}\)

\(P=x^2\)

........

mk chỉ biết làm rút gọn thôi nha

a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)

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

b: P=1/4

=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)

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

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

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

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

...................... 

tìm giá trị x nguyên để A nguyên đi

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. 

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

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