2/\(A=75\left(4^{2004}+4^{2003}+...+1\right)+25\)
\(=25\left(3\left(4^{2004}+4^{2003}+...+1\right)+1\right)\)
\(=25\left(3\left(4^{2004}+4^{2003}+...4\right)+4\right)\)
\(=25.4\left(3\left(4^{2003}+4^{2002}+...+1\right)+1\right)\)
\(=100\left(3\left(4^{2003}+4^{2002}+...+1\right)+1\right)\)
Vậy \(A⋮100\)
Bài 1:
a) \(xy+3x-y=6\)
\(\Rightarrow x\left(y+3\right)-y=6\)
\(\Rightarrow x\left(y+3\right)-\left(y+3\right)=3\)
\(\Rightarrow\left(x-1\right)\left(y+3\right)=3\)
Ta có bảng sau:
| x - 1 | 1 | 3 | -1 | -3 |
| y + 3 | 3 | 1 | -3 | -1 |
| x | 2 | 4 | 0 | -2 |
| y | 0 | -2 | -6 | -4 |
Vậy cặp số \(\left(x;y\right)\) là \(\left(2;0\right);\left(4;-2\right);\left(0;-6\right);\left(-2;-4\right)\)
c) \(\left|x+3\right|+\left|x+1\right|=3x\)
Mà \(\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Rightarrow3x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+3+x+1=3x\)
\(\Rightarrow x=4\)
Vậy x = 4
Ta có:\(23-y^2\le23\)\(\Rightarrow7\left(x-2004\right)^2\le23\)
\(\Rightarrow\left(x-2004\right)^2\le3\)
\(\Rightarrow\left[\begin{matrix}\left(x-2004\right)^2=0\\\left(x-2004\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x-2004=0\\\left[\begin{matrix}x-2004=1\\x-2004=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=2004\\\left[\begin{matrix}x=2005\\x=2003\end{matrix}\right.\end{matrix}\right.\)
Vậy x=2003 hoặc x=2004 hoặc x=2005
