Bài 1:

a) đk: \(x\ne\pm2\)

b) Ta có:

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

\(A=\left[\frac{1}{2-x}-\frac{3x}{\left(2-x\right)\left(2+x\right)}-\frac{2}{2+x}\right]\div\frac{x^2+4+4-x^2}{\left(2-x\right)\left(2+x\right)}\)

\(A=\frac{2+x-3x-2\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}\div\frac{8}{\left(2-x\right)\left(2+x\right)}\)

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

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

=> đpcm

Bài 2: 

a) đk: \(x\ne\left\{-3;0;3\right\}\)

b) Ta có:

\(B=\left[\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right]\div\frac{3x^2}{x+3}\)

\(B=\left[\frac{-x^2-3x-9}{\left(x+3\right)^2}+\frac{x}{x+3}\right]\cdot\frac{x+3}{3x^2}\)

\(B=\frac{-x^2-3x-9+x\left(x+3\right)}{\left(x+3\right)^2}\cdot\frac{x+3}{3x^2}\)

\(B=\frac{-9}{\left(x+3\right)^2}\cdot\frac{x+3}{3x^2}\)

\(B=-\frac{3}{x\left(x+3\right)}\)

c) Khi B = 1/2 thì: \(-\frac{3}{x\left(x+3\right)}=\frac{1}{2}\)

\(\Leftrightarrow x^2+3x=-6\Leftrightarrow x^2+3x+6=0\)

\(\Leftrightarrow\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{15}{4}=0\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2=-\frac{15}{4}\left(ktm\right)\)

Bài 3: 

a) đk: \(x\ne\pm1\)

b) Ta có:

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

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

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

\(P=\frac{4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\cdot\frac{\left(1-x\right)}{4\left(x+1\right)}\)

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

c) Ta có: \(P=\frac{x}{x+1}=1-\frac{1}{x+1}\)để P nguyên thì

\(\frac{1}{x+1}\inℤ\Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\Rightarrow x\in\left\{0;-2\right\}\)

Bài 4:

a) đk: \(\hept{\begin{cases}x\ne\pm y\\x,y\ne0\end{cases}}\)

b) Ta có:

\(Q=\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\left(\frac{x^2}{y}-\frac{y^2}{x}\right)\right]\div\frac{x-y}{x}\)

\(Q=\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\cdot\frac{x^3-y^3}{xy}\right]\div\frac{x-y}{x}\)

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

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

\(Q=\frac{x+y}{y}-\frac{x^2+xy+y^2}{y\left(x+y\right)}\)

\(Q=\frac{\left(x+y\right)^2-x^2-xy-y^2}{y\left(x+y\right)}=\frac{xy}{y\left(x+y\right)}=\frac{x}{x+y}\)

c) Tại \(\hept{\begin{cases}x=-1\\y=1\frac{1}{2}\end{cases}}\) thì: \(Q=\frac{-1}{-1+1\frac{1}{2}}=-2\)

Bài 5:

a) đk: \(\hept{\begin{cases}x\ne1\\y\ne2\end{cases}}\)

b) Ta có:

\(M=\left(\frac{2-x}{x-1}+\frac{4y-4}{y-2}\right)\div\left(\frac{x^2y-x^2}{x-1}+\frac{2y^2-xy^2}{y-2}\right)\)

\(M=\frac{\left(2-x\right)\left(y-2\right)+\left(4y-4\right)\left(x-1\right)}{\left(x-1\right)\left(y-2\right)}\div\frac{\left(x^2y-x^2\right)\left(y-2\right)+\left(2y^2-xy^2\right)\left(x-1\right)}{\left(x-1\right)\left(y-2\right)}\)

\(M=\frac{3xy-2x-2y}{\left(x-1\right)\left(y-2\right)}\cdot\frac{\left(x-1\right)\left(y-2\right)}{3xy^2-3x^2y+2x^2-2y^2}\)

\(M=\frac{3xy-2x-2y}{-3xy\left(x-y\right)+2\left(x-y\right)\left(x+y\right)}\)

\(M=\frac{3xy-2x-2y}{\left(2x+2y-3xy\right)\left(x-y\right)}=\frac{1}{y-x}\)