Để `P=3` thì:
\(\dfrac{2-x}{x^2}=3\\ \Leftrightarrow3x^2-2+x=0\\ \Leftrightarrow3x^2+3x-2x-2=0\\ \Leftrightarrow3x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy ........
Ta có: \(P=3\Leftrightarrow\dfrac{2-x}{x^2}=3\)
\(\Leftrightarrow2-x=3x^2\)
\(\Leftrightarrow3x^2+x-2=0\)
\(\Leftrightarrow3x^2-2x+3x-2=0\)
\(\Leftrightarrow x\left(3x-2\right)+\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2}{3}\end{matrix}\right.\) (tm)
#Urushi
P=3
=>3x^2=2-x
=>3x^2+x-2=0
=>3x^2+3x-2x-2=0
=>(x+1)(3x-2)=0
=>x=-1 hoặc x=2/3
