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ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
a: \(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x\left(x^2+1\right)}{x\left(x+1\right)}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x^2+1\right)}{x+1}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)
Để R=0 thì \(x^2+1=0\)(vô lý)
b: Ta có: |x|=1
=>x=1(loại) hoặc x=-1(loại)
Lời giải:
a) ĐK: $x\neq \pm 2$
b)
\(P=\left[\frac{x^2+2x+4-(x-2)(x+1)}{(x-2)(x^2+2x+4)}-\frac{3}{(x-2)(x^2+2x+4)}\right].\frac{x^2+2x+4}{x^2-4}\)
\(=\frac{3x+6-3}{(x-2)(x^2+2x+4)}.\frac{x^2+2x+4}{(x-2)(x+2)}=\frac{3x+3}{(x+2)(x-2)^2}\)
c)
Để $P$ nhận giá trị dương thì $\frac{3(x+1)}{(x+2)(x-2)^2}>0$. Mà $(x-2)^2>0$ với $x\neq \pm 2$ nên cần tìm $x$ để $\frac{3(x+1)}{x+2}>0$
\(\Rightarrow \left[\begin{matrix} \left\{\begin{matrix} 3(x+1)>0\\ x+2>0\end{matrix}\right.\\ \left\{\begin{matrix} 3(x+1)< 0\\ x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>-1\\ x>-2\end{matrix}\right.\\ \left\{\begin{matrix} x< -1\\ x< -2\end{matrix}\right.\end{matrix}\right.\) hay \(\left[\begin{matrix} x>-1\\ x< -2\end{matrix}\right.\)
Vậy $x>-1; x\neq 2$ hoặc $x< -2$
a, \(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b, \(R=\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
\(=\left(\frac{x^2-2x+1}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
\(=\left(\frac{\left(x^2-2x+1\right)\left(x-1\right)-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\right)\)
\(=\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
\(=\frac{x^3-1}{x^3-1}.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
\(b,\) Để R = 0
\(\Leftrightarrow\frac{x^2+1}{x+1}=0\Leftrightarrow x^2+1=0\) ( vô lý)
Vậy ko có giá trị nào của x để R =0
\(c,\left|R\right|=1\Leftrightarrow\left[{}\begin{matrix}R=-1\\R=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x^2+1}{x+1}=-1\\\frac{x^2+1}{x+1}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=-x-1\\x^2+1=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x^2+x+1\ne0\end{cases}}\)
a/ \(R=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right]\)
\(=1:\left[\frac{x^2+2+\left(x+1\right)\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x}{x^2+x+1}\right)\)
\(=\frac{x^2+x+1}{x}\)
b/ Ta có: \(R=\frac{x^2+x+1}{x}=3+\frac{\left(x-1\right)^2}{x}>3\)
Vậy R > 3