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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
\(c,\left|R\right|=1\Leftrightarrow\left[{}\begin{matrix}R=-1\\R=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x^2+1}{x+1}=-1\\\frac{x^2+1}{x+1}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=-x-1\\x^2+1=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Lời giải:
a) ĐKXĐ: $x\neq 0; x\neq \pm 2$
\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)
\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)
b)
Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)
\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)
\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)
Lời giải:
a) ĐKXĐ: $x\neq 0; x\neq \pm 2$
\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)
\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)
b)
Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)
\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)
\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
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)