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\(P=\left(\dfrac{3x^2+3x-3}{x^2+x-2}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right):\dfrac{1}{x^2-1}\left(dk:x\ne-2,x\ne\pm1\right)\)
\(=\left(\dfrac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right).\left(x^2-1\right)\)
\(=\left(\dfrac{3x^2+3x-3+x+2+x-1-2\left(x^2+x-2\right)}{\left(x-1\right)\left(x+2\right)}\right).\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{3x^2+5x-2-2x^2-2x+4}{x+2}.\left(x+1\right)\\ =\dfrac{x^2+3x+2}{x+2}.\left(x+1\right)\)
\(=\dfrac{x^2+x+2x+2}{x+2}.\left(x+1\right)\\ =\dfrac{x\left(x+1\right)+2\left(x+1\right)}{x+2}.\left(x+1\right)\\ =\dfrac{\left(x+1\right)^2\left(x+2\right)}{x+2}\\ =x^2+2x+1\)
Ta có :
\(x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Với \(x=3\) thì \(P=x^2+2x+1=\left(x+1\right)^2=\left(3+1\right)^2=16\)
Vậy ...
a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)
b: A>0
=>x+1>0
=>x>-1
c: x^2+3x+2=0
=>(x+1)(x+2)=0
=>x=-2(loại) hoặc x=-1(loại)
Do đó: Khi x^2+3x+2=0 thì A ko có giá trị
B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)
\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)
b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)
\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))
\(\Leftrightarrow x>-1\).
-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).
a: \(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2-1}{2x+1}\)
\(=\dfrac{2}{2x+1}\)
b: Để \(P=\dfrac{3}{x-1}\) thì \(\dfrac{3}{x-1}=\dfrac{2}{2x+1}\)
=>6x+3=2x-2
=>4x=-5
hay x=-5/4
\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)
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: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
Bạn xem lại \(a,b\) mình làm rồi nha.
\(c,P>0\Leftrightarrow\left(x+1\right)^2>0\) (luôn đúng \(\forall x\))
Vậy với mọi giá trị x thì \(P>0\).