\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\sqrt{\dfrac{a^2\left(a+1\right)^2+a^2+\left(a+1\right)^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)
\(=\sqrt{\dfrac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)
\(=\sqrt{\dfrac{\left(a\left(a+1\right)+1\right)^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}=\dfrac{a\left(a+1\right)+1}{a+1}+\dfrac{a}{a+1}\)
\(=\dfrac{a^2+2a+1}{a+1}=\dfrac{\left(a+1\right)^2}{a+1}=a+1\)
\(\Rightarrow VP=6831\)
Không làm mất tính tổng quát, giả sử \(x\le y\le z\)
Dễ dàng kiểm chứng \(x=y=z\) không phải là nghiệm
\(3^x+3^y+3^z=6831\Leftrightarrow3^x\left(1+3^{y-x}+3^{z-x}\right)=3^3.253\)
Nếu \(1+3^{y-x}+3^{z-x}\ne253\Rightarrow1+3^{y-x}+3^{z-x}=253.3^k⋮3\)
Nhưng \(1+3^{y-x}+3^{z-x}⋮̸3\) với \(\left\{{}\begin{matrix}x\ne y\\x\ne z\end{matrix}\right.\)\(\Rightarrow\) vô lý
Vậy \(\left\{{}\begin{matrix}3^x=3^3\\1+3^{y-x}+3^{z-x}=253\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\3^{y-3}+3^{z-3}=252\end{matrix}\right.\)
\(\Rightarrow3^{y-3}\left(1+3^{z-y}\right)=252=3^2.28\)
Do \(3^{z-y}+1⋮̸3\) lý luậnt ương tự như trên \(\Rightarrow\left\{{}\begin{matrix}3^{y-3}=3^2\\1+3^{z-y}=28\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y-3=2\\3^{z-y}=27=3^3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=5\\z=8\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=3\\y=5\\z=8\end{matrix}\right.\)
