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29 tháng 4 2018

a)  \(ĐKXĐ:\) \(x\ne\pm1\)

\(A=\left(\frac{3x^2-4}{x^2-1}-\frac{2}{1-x}-\frac{2}{x+1}\right):\left(\frac{1-x}{x+1}\right)\)

\(=\left(\frac{3x^2-4}{\left(x-1\right)\left(x+1\right)}+\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+1}{1-x}\)

\(=\frac{3x^2-4+2x+2-2x+2}{\left(x-1\right)\left(x+1\right)}.\frac{x+1}{1-x}\)

\(=\frac{3x^2}{\left(x-1\right)\left(x+1\right)}.\frac{x+1}{1-x}\)

\(=-\frac{3x^2}{\left(x-1\right)^2}\)

1 tháng 12 2021

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

a: \(A=\dfrac{3x^2-4+2\left(x+1\right)-2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x+1}{x+1-x}\)

\(=\dfrac{3x^2-4+2x+2-2x+2}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x+1}{1}\)

\(=\dfrac{3x^2}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{x+1}{1}=\dfrac{3x^2}{x-1}\)

b: Để A chia hết cho 2013 thì A=2013k

=>3x2=2013k(x-1)(k∈Z)