-1/14 nha bn 

Để cho \(P\left(x\right)\) có nghiệm thì \(7x+\dfrac{1}{2}=0\)

\(\Rightarrow7x=-\dfrac{1}{2}\)

\(\Rightarrow x=-\dfrac{1}{14}\)

\(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=x+y+z\left(1\right)\)

Ta có: 

\(\dfrac{y+z+1}{x}=\dfrac{z+x+z}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+z+x+2+x+y-3}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}\left(2\right)\)

Trường hợp 1: \(x+y+z=0\Leftrightarrow x=y=z=0\)

Trường hợp 2: \(x+y+z\ne0\)

\(\left(2\right)\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+z=3x-1\\x+y+z=3y-2\\x+y+z=3z+3\end{matrix}\right.\Rightarrow3x-1=3y-2=3z+3\Rightarrow\left\{{}\begin{matrix}x=z+\dfrac{4}{3}\\y=z+\dfrac{5}{3}\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\dfrac{z}{z+\dfrac{4}{3}+z+\dfrac{5}{3}-3}=z+\dfrac{4}{4}+z+\dfrac{5}{4}+z\)

     \(\Rightarrow\dfrac{1}{2}=3+3z\)

     \(\Rightarrow z=-\dfrac{5}{6}\)

\(\Rightarrow x=-\dfrac{5}{6}+\dfrac{4}{3}=\dfrac{1}{2}\)

\(\Rightarrow y=-\dfrac{5}{6}+\dfrac{5}{3}=\dfrac{5}{6}\)

$8<2^n<64$$\iff 2^3<2^n<2^6$$\iff 3<n<6$

Vì n nguyên dương $\Rightarrow n\in \left\{4;5\right\}$

Vậy $n\in \left\{4;5\right\}$

$\_HT\_$