a: ĐKXĐ của A là \(x\notin\left\{0;\dfrac{1}{2}\right\}\)
ĐKXĐ của B là \(x\ne1\)
b: \(A=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{-3x+1}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(-3x+1\right)\left(2x-1\right)+2x\left(3x-2\right)-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-6x^2+3x+2x-1+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}=\dfrac{-1}{2x}\)
\(B=\dfrac{6x^2+8x+7}{x^3-1}+\dfrac{x}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{6x^2+8x+7}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(=\dfrac{6x^2+8x+7+x\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{6x^2+8x+7+x^2-x-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x-1}\)
c: Để A>0 và B<0 thì \(\left\{{}\begin{matrix}-\dfrac{1}{2x}>0\\\dfrac{1}{x-1}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{2x}< 0\\x-1< 0\end{matrix}\right.\Leftrightarrow x< 0\)
Kết hợp ĐKXĐ, ta được:
x<0
