a/ \(\left\{{}\begin{matrix}xy=2016\\x+y=-95\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-95\\\left(-y-95\right)y=2016\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-95\\y^2+95y+2016=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-95\\\left(y^2+32y\right)+\left(63y+2016\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-95\\y\left(y+32\right)+63\left(y+32\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-95\\\left(y+32\right)\left(y+63\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-32\\x=-63\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}y=-63\\x=-32\end{matrix}\right.\)
c/ Vì x nguyên dương nên dễ thấy
\(7^y=x^3+5x^2+21>x+5=7^z\)
\(\Leftrightarrow y>z\)
Xét \(y>z>1\)
Ta có:
\(7^y=x^3+5x^2+21=x^2.7^z+21\)
\(\Leftrightarrow7^{y-1}-x^2.7^{z-1}=3\) không thỏa mãn vì vế trái chia hết cho 7 VP không chia hết cho 7.
Xét \(z=1\)
\(\Rightarrow x=7^1-5=2\)
\(\Rightarrow7^y=2^3+5.2^2+21=49=7^2\)
\(\Rightarrow y=2\)
Vậy giá trị x, y, z cần tìm là: (x; y; z) = (2; 2; 1)