a) ĐKXĐ: \(x\notin\left\{1;\frac{3}{2}\right\}\)
Ta có: \(P=\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
\(=\left(\frac{2x}{\left(x-1\right)\left(2x-3\right)}-\frac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):\left(\frac{3\left(x-1\right)\left(2x-3\right)}{\left(x-1\right)\left(2x-3\right)}-\frac{2\left(2x-3\right)}{\left(x-1\right)\left(2x-3\right)}\right)\)
\(=\frac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{3\left(2x^2-5x+3\right)-2\left(2x-3\right)}{\left(x-1\right)\left(2x-3\right)}\)
\(=\frac{-3x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{6x^2-15x+9-4x+6}{\left(x-1\right)\left(2x-3\right)}\)
\(=\frac{-3x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{6x^2-19x+15}{\left(x-1\right)\left(2x-3\right)}\)
\(=\frac{-\left(3x-5\right)}{\left(2x-3\right)\left(x-1\right)}\cdot\frac{\left(x-1\right)\left(2x-3\right)}{\left(2x-3\right)\left(3x-5\right)}\)
\(=\frac{-1}{2x-3}\)
b) Ta có: \(\left|3x-2\right|+1=5\)
\(\Leftrightarrow\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{2}{3}\end{matrix}\right.\)
Thay x=3 vào biểu thức \(P=\frac{-1}{2x-3}\), ta được:
\(\frac{-1}{2\cdot3-3}=\frac{-1}{6-3}=\frac{-1}{3}\)
Thay \(x=-\frac{2}{3}\) vào biểu thức \(P=\frac{-1}{2x-3}\), ta được:
\(-\frac{1}{2\cdot\frac{-2}{3}-3}=-\frac{1}{\frac{-4}{3}-\frac{9}{3}}\)
\(=-1:\frac{-13}{3}=-1\cdot\frac{3}{-13}=\frac{-3}{-13}=\frac{3}{13}\)
Vậy: Khi |3x-2|+1=5 thì \(P\in\left\{-\frac{1}{3};\frac{3}{13}\right\}\)
c) Để P>0 thì \(\frac{-1}{2x-3}>0\)
\(\Leftrightarrow-1;2x-3\) cùng dấu
mà -1<0
nên 2x-3<0
\(\Leftrightarrow2x< 3\)
hay \(x< \frac{3}{2}\)
mà \(x\notin\left\{1;\frac{3}{2}\right\}\)(ĐKXĐ của P)
nên \(\left\{{}\begin{matrix}x< \frac{3}{2}\\x\ne1\end{matrix}\right.\)
Vậy: Khi \(\left\{{}\begin{matrix}x< \frac{3}{2}\\x\ne1\end{matrix}\right.\) thì P>0
d) Để \(P=\frac{1}{6-x^2}\) thì \(\frac{-1}{2x-3}=\frac{1}{6-x^2}\)
\(\Leftrightarrow-1\cdot\left(6-x^2\right)=1\left(2x-3\right)\)
\(\Leftrightarrow x^2-6=2x-3\)
\(\Leftrightarrow x^2-6-2x+3=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: Khi \(P=\frac{1}{6-x^2}\) thì \(x\in\left\{3;-1\right\}\)