\(\frac{x+1}{3x}-x-1\)

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

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

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

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

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

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

SO₂:  2,2   ---    0,2   ----  4.48lít

NH₃:   0,6  ----  12,75

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x-14}=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}\)

\(=\frac{1}{x-3}\)

Phân thức xác định

\(\Leftrightarrow2x^2-2\ne0\)

\(\Leftrightarrow2\left(x^2-1\right)\ne0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\ne0\)

\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

Vậy phân thức xác định \(\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

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

Thay x=-2 vào A ta có: \(A=\frac{2}{-2+1}=\frac{2}{-1}=-2\)

Vậy \(A=-2\)tại x=-2

Ta có: \(x\in Z\Rightarrow x+1\in Z\)

\(A\in Z\Leftrightarrow\left(x+1\right)\in\text{Ư}\left(2\right)=\left\{\pm1;\pm2\right\}\)

đến đây b tự làm nhé~

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5.\left(x+1\right)}\)

\(A=\left(\frac{x^2+2x+1}{\left(x+1\right).\left(x-1\right)}-\frac{x^2-2x+1}{\left(x+1\right).\left(x-1\right)}\right):\frac{2x}{5.\left(x+1\right)}\)

\(A=\frac{x^2+2x+1-x+2x-1}{\left(x+1\right).\left(x-1\right)}\cdot\frac{5.\left(x+1\right)}{2x}\)

\(A=\frac{4x}{\left(x+1\right).\left(x-1\right)}\cdot\frac{5.\left(x+1\right)}{2x}=\frac{10}{x-1}\)

Cảm ơn bn nhiều!