\(ĐKXĐ:x\ne-1\)

\(Q=\frac{2x^2+2}{\left(x+1\right)^2}=\frac{\left(x^2-2x+1\right)+\left(x^2+2x+1\right)}{\left(x+1\right)^2}=\frac{\left(x-1\right)^2+\left(x+1\right)^2}{\left(x+1\right)^2}=\frac{\left(x-1\right)^2}{\left(x+1\right)^2}+1\ge1\forall x\)

Dấu "=" xảy ra khi x - 1 = 0 tức x = 1

Vậy GTNN của Q là 1 khi x = 1

ta có: 2x² - x + 1 chia hết cho 2x + 1

2x² + x - 2x - 1 + 2 chia hết cho 2x + 1

x.(2x+1) - (2x+1) + 2 chia hết cho 2x + 1

(x-1).(2x+1) + 2 chia hết cho 2x + 1

mà (x-1).(2x+1) chia hết cho 2x + 1

=> 2 chia hết cho 2x + 1

=> ...

bn tự làm tiếp nha

\(4x\left(x-2007\right)-x+2007=0\)

\(4x\left(x-2007\right)-\left(x-2007\right)=0\)

\(\left(x-2007\right)\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2007=0\\4x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2007\\x=\frac{1}{4}\end{cases}}\)

Vậy....

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

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

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

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

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

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

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

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

NH₃:   0,6  ----  12,75

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

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

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

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

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