Answer:

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

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

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

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

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

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

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=\frac{-5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}Q=\frac{4}{3}\\Q=\frac{1}{2}\end{cases}}\)

Answer:

Để cho phân thức F được xác định thì

\(x^3-8\ne0\)

\(\Leftrightarrow x^3\ne8\)

\(\Leftrightarrow x\ne2\)

Điều kiện xác định của \(A\)là: 

\(\hept{\begin{cases}2-x\ne0\\4-x^2\ne0\\2+x\ne0\end{cases}}\Leftrightarrow x\ne\pm2\).

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

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

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

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

\(A=-5\Rightarrow\frac{4x}{2-x}=-5\Rightarrow4x=5\left(x-2\right)\Leftrightarrow x=10\)(thỏa mãn) 

\(x^2-4x+y+4\)

\(=\left(x^2-4x+4\right)+y\)

\(=\left(x-2\right)^2+y\)