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

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

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

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

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

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

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

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

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

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

Xin phép sửa đề:

Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}=\frac{x+3}{1-x^2}\) \(\left(x\ne\pm1\right)\)

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

\(\Rightarrow3x^2+4x+1-1+2x-x^2=-x^2-2x+3\)

\(\Leftrightarrow3x^2+8x-3=0\)

\(\Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

Vậy tập nghiệm PT \(S=\left(-3;\frac{1}{3}\right)\)

\(A=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})(1-\frac{1}{1+2+3+4})...(1-\frac{1}{1+2+3+...+2006})\)

\(A=(1-\frac{1}{3})(1-\frac{1}{6})(1-\frac{1}{10})...(1-\frac{1}{2013021})\)

\(A=\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}....\frac{2013020}{2013021}\)

Sorry bạn máy tính mình có chút vấn đề để mk làm tiếp :

\(A=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}....\cdot\frac{4026040}{4026042}\)

\(A=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{2005\cdot2008}{2006\cdot2007}\)

\(A=\frac{1\cdot2\cdot3\cdot...\cdot2005}{2\cdot3\cdot4\cdot...\cdot2006}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2008}{3\cdot4\cdot5\cdot...\cdot2007}\)

\(A=\frac{1}{2006}\cdot\frac{2008}{3}=\frac{1004}{3009}\)

P/S : Hoq chắc :>

dsf