a)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)

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

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

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

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

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

b)Bài này liên quan đến dấu lớn nên mk ko làm đc

Để x thuộc Z thì : à dấu " : " là " chia hết cho " nhá ^^

2x - 1 : x + 2

2x + 2 -3 : x + 2

mà 2x + 2 : x + 2 => 3 : x + 2 => x + 2 thuộc Ư(3) = { 1; -1; 3; -3 }

+) x + 2 = 1

x = -1

+) x + 2 = -1

x = -3

+) x + 2 = 3

x = 1

+) x + 2 = -3

x = -5

Vậy,.........

Để x thuộc Z thì : à dấu " : " là " chia hết cho " nhá ^^
2x - 1 : x + 2
2x + 2 -3 : x + 2
mà 2x + 2 : x + 2 => 3 : x + 2 => x + 2 thuộc Ư(3) = { 1; -1; 3; -3 }
+) x + 2 = 1
x = -1
+) x + 2 = -1
x = -3
+) x + 2 = 3
x = 1
+) x + 2 = -3
x = -5
Vậy,.........

\(ĐKXĐ:x\ne2;x\ne-2;x\ne0\)

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

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

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

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

\(b,P=\frac{-4}{2+x}=\frac{1}{2}\)

\(2+x=-8\)

\(x=-10\)

\(c,P=-\frac{4}{2+x}\)

\(< =>-4⋮x+2\)

lập bảng ra thì bạn ra đc \(x=\left\{2;-1;-3;-6\right\}\)

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

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

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

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

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

b) Để P=1/2

\(\Rightarrow-\frac{4}{x+2}=\frac{1}{2}\)

\(\Leftrightarrow-8=x+2\)

\(\Leftrightarrow x=-10\)

c) Để P nhận GT nguyên

\(\Rightarrow\left(x+2\right)\inƯ_{\left(-4\right)}\)

\(\Rightarrow\left(x+2\right)\in\left\{-1;1;-2;2;-4;4\right\}\)

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

#H

Của bạn đây nhé 

\(a,ĐKXĐ:x\ne0;x\ne1\)

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

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

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

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

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

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