a) \(B=\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right)\div\left(3+\frac{2}{1-x}\right)\)
\(=\left(\frac{2x}{\left(2x-3\right)\left(x-1\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right)\div\frac{3\left(1-x\right)+2}{1-x}\)
\(=\frac{2x-\left(5x-5\right)}{\left(2x-3\right)\left(x-1\right)}.\frac{1-x}{5-3x}\)\(=\frac{\left(5-3x\right)\left(1-x\right)}{\left(2x-3\right)\left(x-1\right)\left(5-3x\right)}=\frac{1}{3-2x}\)
b) Để \(B=\frac{1}{x^2}\) thì :
\(\frac{1}{3-2x}=\frac{1}{x^2}\Leftrightarrow3-2x=x^2\)
\(\Leftrightarrow x^2+2x-3=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
c) Để B > 0 thì :
\(\frac{1}{3-2x}>0\Leftrightarrow3-2x>0\)
\(\Leftrightarrow2x< 3\Leftrightarrow x< \frac{3}{2}\)
