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

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

\(\Rightarrow3x=2\left(x^2-x+1\right)\)

\(\Leftrightarrow2x^2-2x+2-3x=0\)

\(\Leftrightarrow2x^2-5x+2=0\)

\(\Leftrightarrow2x^2-4x-x+2=0\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)

Với x = 2 => Q = 4/21

Với x = 1/2 => Q = 4/21 :))

"Trần Nhật Quỳnh" có cách này ngắn gọn hơn nữa.

Ta có: 

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

\(\Rightarrow x+\frac{1}{x}=\frac{5}{2}\)

Lại có:

\(Q=\frac{x^2}{x^4+x^2+1}\)

\(\frac{1}{Q}=\frac{x^4+x^2+1}{x^2}\)

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

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

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

Vì \(x+\frac{1}{x}=\frac{5}{2}\)nên

\(\frac{1}{Q}=\left(\frac{5}{2}\right)^2-2\)

\(\frac{1}{Q}=\frac{25}{4}-2\)

\(\frac{1}{Q}=\frac{21}{4}\)

\(\Rightarrow Q=\frac{4}{21}\)

Vậy \(Q=\frac{4}{21}\)

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!!

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

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

Bài làm

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

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

(bước trên là mình đổi dấu ở phân số thứ hai, dấu âm chuyển xuống dưới mẫu nên đổi dấu ở mẫu, sau đó nhân với cả cụm x + 1 nha, tại hơi tắt nên thêm dòng giải thích cho dễ hiểu)

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

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

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

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

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

=> \(x-\frac{3}{4}=\pm\frac{5}{4}\)

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

*Trường hợp 1: Khi x = 2

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

\(Q=\frac{2^2}{2+1}=\frac{4}{3}\)

Vậy khi x = 2 thì Q = 4/3

*Trường hợp 2: Khi x = -1/2

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

\(Q=\frac{\left(-\frac{1}{2}\right)^2}{-\frac{1}{2}+1}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}:\frac{1}{2}=\frac{1}{4}\cdot2=\frac{1}{2}\)

Vậy x = -1/2 thì Q = 1/2