a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)
b: \(Q=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}-\dfrac{3x+1-x^2}{3x}\)
\(=\left(\dfrac{\left(x+2\right)\left(x+1\right)+2\cdot3x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right)\cdot\dfrac{x+1}{2\left(1-2x\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2\left(1-2x\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{2\left(1-2x\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{2\left(4x^2-1\right)}{3x\cdot2\cdot\left(2x-1\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{3x\left(2x-1\right)}+\dfrac{x^2-3x-1}{3x}\)
\(=\dfrac{2x+1}{3x}+\dfrac{x^2-3x-1}{3x}=\dfrac{x^2-x}{3x}=\dfrac{x-1}{3}\)
c: Để Q<0 thì \(\dfrac{x-1}{3}< 0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-1;\dfrac{1}{2}\right\}\end{matrix}\right.\)
d: \(P\cdot Q=\dfrac{\left(x-1\right)}{3}\cdot\left(x-2\right)=\dfrac{1}{3}\left(x^2-3x+2\right)\)
\(=\dfrac{1}{3}\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{12}>=-\dfrac{1}{12}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=3/2
`#3107.101107`
`a)`
ĐKXĐ của Q: \(\left\{{}\begin{matrix}3x\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
`b)`
\(Q=\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right)\div\dfrac{2-4x}{x+1}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2-4x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{2-8x^2}{3x}\cdot\dfrac{1}{2-4x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{2\left(1-4x^2\right)}{3x\left(2-4x\right)}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{2\left(1-2x\right)\left(1+2x\right)}{3x\cdot2\left(1-2x\right)}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{1+2x}{3x}-\dfrac{3x+1-x^2}{3x}\)
\(=\dfrac{1+2x-\left(3x+1-x^2\right)}{3x}\)
\(=\dfrac{1+2x-3x-1+x^2}{3x}\)
\(=\dfrac{x^2-x}{3x}=\dfrac{x\left(x-1\right)}{3x}=\dfrac{x-1}{3}\)
`c)`
Để `Q < 0:`
\(\Rightarrow\dfrac{x-1}{3}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)
Theo ĐKXĐ: \(x\notin\left\{-1;0;\dfrac{1}{2}\right\}\)
`d)`
`P = x - 2`
\(\Rightarrow PQ=\left(x-2\right)\left(\dfrac{x-1}{3}\right)=\dfrac{1}{3}\left(x-2\right)\left(x-1\right)=\dfrac{1}{3}\left(x^2-3x+2\right)\)
\(=\dfrac{1}{3}\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\right)\\ =\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{12}\)
`=> PQ = 1/3(x-3/2)^2 - 1/12 \ge - 1/12` (Thỏa mãn ĐKXĐ)
`=>` Dấu `"="` xảy ra khi: `(x - 3/2)^2 = 0 => x = 3/2.`
