1) \(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Rightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Rightarrow x^2-2x+1+\left(2y\right)^2-2.2y.3+9+z^2+2.z.2+4=0\)
\(\Rightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0\) với mọi x
\(\left(2y-3\right)^2\ge0\) với mọi y
\(\left(z+2\right)^2\ge0\) với mọi z
Mà \(\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(2y-3\right)^2=0\\\left(z+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\2y=3\\z=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
Vậy x = 1 ; y = 3/2 ; z = -2
2) a)
Ta có:
\(103n^2+121n+70\)
\(=103n^2-103n+224n-224+294\)
\(=103n\left(n-1\right)+224\left(n-1\right)+294\)
\(=\left(n-1\right)\left(103n+224\right)+294\)
Vì ( n - 1 )( 103n + 224 ) chia hết cho n - 1
=> Để 103n2 + 121n + 70 chia hết cho n - 1
=> 294 phải chia hết cho n - 1
=> n - 1 thuộc Ư(294)
=> n - 1 thuộc { 2 ; -2 ; 3 ; -3 ; 7 ; -7 ; 49 ; -49 ; 6 ; - 6 ; 21 ; -21 ; 147 ; -147 ; 14 ; -14 ; 98 ; -98 ; 1 ; -1 ; 294 ; -294 }
=> n thuộc { 3 ; -1 ; 4 ; -2 ; 8 ; -6 ; 50 ; -48 ; 7 ; -5 ; 22 ; -20 ; 148 ; -146 ; 15 ; -13 ; 99 ; -97 ; 2 ; 0 ; 295 ; -293 }
