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TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a: ĐKXĐ: x<>1; x<>2; x<>3
\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)
\(=\dfrac{x^3-x^2+x^3-3x^2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}\)
\(=\dfrac{2x^3-4x^2}{\left(x-2\right)}\cdot\dfrac{1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\dfrac{2x^2}{x^4+x^2+1}\)
b:
1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)
2: \(=\dfrac{3x+1-2x^2-2x+3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)
\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)
b: x=2 ko thỏa mãn ĐKXĐ
=>Loại
Khi x=3 thì A=-1/(3-2)=-1
c: A=2
=>x-2=-1/2
=>x=3/2
\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)
a: \(P=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{x-1}\)
\(=\dfrac{1-1-3x}{3x}\cdot\dfrac{2x}{x-1}\)
\(=\dfrac{-3x}{3x}\cdot\dfrac{2x}{x-1}=\dfrac{-2x}{x-1}\)
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.`