a) ĐKXĐ: \(x\notin\left\{1;\frac{3}{2}\right\}\)
Ta có: \(B=\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
\(=\left(\frac{2x}{\left(x-1\right)\left(2x-3\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):\left(\frac{3\left(x-1\right)}{x-1}-\frac{2}{x-1}\right)\)
\(=\frac{2x-5x+5}{\left(x-1\right)\left(2x-3\right)}:\frac{3x-5}{x-1}\)
\(=\frac{-3x+5}{\left(x-1\right)\left(2x-3\right)}\cdot\frac{x-1}{3x-5}\)
\(=\frac{-\left(3x-5\right)\cdot\left(x-1\right)}{\left(x-1\right)\cdot\left(2x-3\right)\cdot\left(3x-5\right)}=\frac{-1}{2x-3}\)
b) Ta có: \(B=\frac{1}{x^2}\)
⇔\(\frac{-1}{2x-3}=\frac{1}{x^2}\)
⇔\(x^2+2x-3=0\)
\(\Leftrightarrow x^2+3x-x-3=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy: Khi x=-3 thì \(B=\frac{1}{x^2}\)
c) Để B>0 thì \(\frac{-1}{2x-3}>0\)
mà -1<0
nên 2x-3<0
⇔2x<3
hay \(x< \frac{3}{2}\)
Vậy: Khi \(x< \frac{3}{2}\) thì B>0
