Question

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

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RÚt gọn

Answer 1

Em thử nhá

ĐKXĐ: \(\left\{{}\begin{matrix}x^3-1\ne0\\1-x\ne0\\2x^2+x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\frac{1}{2};x\ne-1\end{matrix}\right.\Leftrightarrow x\ne\left\{-1;1;\frac{1}{2}\right\}\)

Rút gọn: \(B=\left[\frac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+3}{-\left(1-x\right)}\right].\frac{x-1}{\left(2x-1\right)\left(x+1\right)}\)

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

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

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