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

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

a) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)

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

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

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

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

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

b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:

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

*TH1: 

\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)

*TH2:

\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)

\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)

c,

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

Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0;x\ne2\\x\ne-1\end{cases}}\)

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

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

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

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

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

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

\(\Leftrightarrow Q=1+\frac{-2}{x+1}\)

\(\Leftrightarrow Q=\frac{x-1}{x+1}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(ktm\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)

Thay \(x=-\frac{1}{2}\)vào Q, ta được :

\(Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}\)

\(\Leftrightarrow Q=\frac{-\frac{3}{2}}{\frac{1}{2}}\)

\(\Leftrightarrow Q=-3\)

c) Để \(Q\inℤ\)

\(\Leftrightarrow x-1⋮x+1\)

\(\Leftrightarrow x+1-2⋮x+1\)

\(\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

đkxd: \(x\ne\left\{\pm3\right\}\)

a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)

=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)

=\(\frac{x+12}{x-3}\)

b)|2x+1|=5

<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2

với x=-3 thì B=\(\frac{-3}{2}\)

với x=2 thì B=-14