\(a,DKXD:x\ne0;3\)\(b,\) Thay \(x=3\) vào P \(\Leftrightarrow\dfrac{2.3+3}{3\left(3-3\right)}+\dfrac{1}{3}\)\(\Leftrightarrow P=\dfrac{1}{3}\)Vậy \(P=\dfrac{1}{3}\) khi \(x=3\)\(c,\) Thay \(P=\dfrac{1}{2}\) vào bểu thức\(\Leftrightarrow\dfrac{1}{2}=\dfrac{2x+3}{x\left(x-3\right)}+\dfrac{1}{x}\)\(\Leftrightarrow\dfrac{2x+3+x-3}{x\left(x-3\right)}-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{2x+3+x-3-x^2+3x}{2x\left(x-3\right)}=0\)\(\Leftrightarrow-x^2=0\Leftrightarrow x=0\)Vậy \(x=0\) để \(P=\dfrac{1}{2}\)Câu trả lời: \(a,DKXD:x\ne0;3\)\(b,\) Thay \(x=3\) vào P \(\Leftrightarrow\dfrac{2.3+3}{3\left(3-3\right)}+\dfrac{1}{3}\)\(\Leftrightarrow P=\dfrac{1}{3}\)Vậy \(P=\dfrac{1}{3}\) khi \(x=3\)\(c,\) Thay \(P=\dfrac{1}{2}\) vào bểu thức\(\Leftrightarrow\dfrac{1}{2}=\dfrac{2x+3}{x\left(x-3\right)}+\dfrac{1}{x}\)\(\Leftrightarrow\dfrac{2x+3+x-3}{x\left(x-3\right)}-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{2x+3+x-3-x^2+3x}{2x\left(x-3\right)}=0\)\(\Leftrightarrow-x^2=0\Leftrightarrow x=0\)Vậy \(x=0\) để \(P=\dfrac{1}{2}\)