Question

Giúp em vs ạ:(

Answer 1

\(a,DKXD:x\ne0;3\)

\(b,\) 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 \(P=\dfrac{1}{2}\)