a/ Trừ pt dưới cho pt đầu: \(\Rightarrow x-z=-1\Rightarrow x=z-1\)
Thay vào pt cuối:
\(\left(3z-1\right)^2=25\Rightarrow\left[{}\begin{matrix}3z-1=5\\3x-1=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}z=2\\z=-\frac{4}{3}\end{matrix}\right.\)
TH1: \(z=2\Rightarrow x=1\Rightarrow y=4-2x=2\)
TH2: \(z=-\frac{4}{3}\Rightarrow x=-\frac{7}{3}\Rightarrow y=\frac{26}{3}\)
b/ Theo Viet đảo, ta có \(x^2\) và \(y^2\) là nghiệm của:
\(t^2-5t+4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=1\\y^2=4\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=4\\y^2=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\pm1\\y=\pm2\end{matrix}\right.\\\left\{{}\begin{matrix}x=\pm2\\y=\pm1\end{matrix}\right.\end{matrix}\right.\)
c/
\(\left(2x^3+x\right)^2=9\Leftrightarrow\left[{}\begin{matrix}2x^3+x=3\\2x^3+x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^3+x-3=0\\2x^3+x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(2x^2+2x+3\right)=0\\\left(x+1\right)\left(2x^2-2x+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
