a: ĐKXĐ: x∉{1;-1;2}

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

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

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

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

b: Để P nguyên thì \(x^2\) ⋮x-2

=>\(x^2-4+4\) ⋮x-2

=>4⋮x-2

=>x-2∈{1;-1;2;-2;4;-4}

=>x∈{3;1;4;0;6;-2}

Kết hợp ĐKXĐ, ta được: x∈{3;4;0;6;-2}

c: \(P=\frac{x^2}{x-2}\)

\(=\frac{x^2-4+4}{x-2}=x+2+\frac{4}{x-2}=x-2+\frac{4}{x-2}+4\ge2\cdot\sqrt{\left(x-2\right)\cdot\frac{4}{x-2}}+4\)

=>P>=2*2+4=8

Dấu '=' xảy ra khi \(\left(x-2\right)^2=4\)

=>x-2=2

=>x=4(nhận)

a: Ta có: \(\frac{x}{x^2-2x+1}+\frac{x}{1-x^2}\)

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

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

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

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

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

a) Giải:

Để A có giá trị là số nguyên thì \(x+1⋮x-2\)

Ta có:
\(x+1⋮x-2\)

\(\Rightarrow\left(x-2\right)+3⋮x-2\)

\(\Rightarrow3⋮x-2\)

\(\Rightarrow x-2\in\left\{\pm1;\pm3\right\}\)

+) \(x-2=1\Rightarrow x=3\)

+) \(x-2=-1\Rightarrow x=1\)

+) \(x-2=3\Rightarrow x=5\)

+) \(x-2=-3\Rightarrow x=-1\)

Vậy \(x\in\left\{3;1;5;-1\right\}\)

b) Để B có giá trị nguyên thì \(2x-1⋮x+5\)

Ta có:

\(2x-1⋮x+5\)

\(\Rightarrow\left(2x+10\right)-9⋮x+5\)

\(\Rightarrow2.\left(x-5\right)-9⋮x+5\)

\(\Rightarrow-9⋮x+5\)

\(\Rightarrow x+5\in\left\{\pm1;\pm3;\pm9\right\}\)

+) \(x+5=1\Rightarrow x=-4\)

+) \(x+5=-1\Rightarrow x=-6\)

+) \(x+5=3\Rightarrow x=-2\)

+) \(x+5=-3\Rightarrow x=-8\)

+) \(x+5=9\Rightarrow x=4\)

+) \(x+5=-9\Rightarrow x=-14\)

Vậy \(x\in\left\{-4;-;-2;-8;4;-14\right\}\)
 

a, Để phân số đạt giá trị nguyễn 

\(\Rightarrow x+1⋮x-2\)

\(\Rightarrow x-2+3⋮x-2\)

mà \(x-2⋮x-2\Rightarrow3⋮x-2\)

\(\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{3;5\pm1\right\}\)

b,Tương tự :

\(2x-1⋮x+5\)

\(\Rightarrow2x+10-11⋮x+5\)

\(2\left(x+5\right)-11⋮x+5\)

mà \(2\left(x+5\right)⋮x+5\Rightarrow11⋮x+5\)

\(\Rightarrow x+5\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

\(x\in\left\{-4;\pm6;-16\right\}\)

a, Để \(A\in Z\)\(\Leftrightarrow x+1⋮x-2\)\

Ta có:              \(\hept{\begin{cases}x-2⋮x-2\\x+1⋮x-2^{ }_{ }\end{cases}}\) \(\Rightarrow\) \(x-2-\left(x+1\right)⋮x-2\)

\(\Rightarrow-3⋮x-2\)mà \(x\in Z\Rightarrow x-2\in Z\Rightarrow x-2\inƯ\left(-3\right)\)\(\in\left(1;-1,3;-3\right)\)

\(x\in\left(3;1;5;-1\right)\)Vậy: \(x\in\left(1;3;5;-1\right)\)thì \(A\in Z\)

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

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

Để A \(\in Z\Leftrightarrow\frac{2}{x^2+x+1}\in Z\Rightarrow x^2+x+1\inƯ\left(2\right)\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\in\left\{\pm1;\pm2\right\}\)

giải ra ta được \(x=0,x=-1\)(t/m)