1/ \(3^{3x+3}=27\)
\(\Leftrightarrow3^{3x+3}=3^3\)
\(\Leftrightarrow3x+3=3\)
\(\Leftrightarrow3x=0\)
\(\Leftrightarrow x=0\)
2/ \(2x+6⋮x+2\)
Mà \(x+2⋮x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+6⋮x+2\\2x+4⋮x+2\end{matrix}\right.\)
\(\Leftrightarrow2⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=1\\x+2=2\\x+2=-1\\x+2=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-3\\x=-4\end{matrix}\right.\)
Vậy ..
\(3^{3x+3}=27\\ \Leftrightarrow3^{3x+3}=3^3\\ 3\ne0;3\ne\pm1\\ \Rightarrow3x+3=3\\ \Leftrightarrow3x=3-3\\ \Leftrightarrow3x=0\\ \Leftrightarrow x=0\)
Vậy ...
\(2x+6=2x+4+2=2\left(x+2\right)+2\\ 2\left(x+2\right)⋮\left(x+2\right)\\ \text{Để }\left(2x+6\right)⋮\left(x+2\right)\Rightarrow2⋮\left(x+2\right)\Rightarrow\left(x+2\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
| $ x + 2 $ | $ x $ |
| $ - 2 $ | $ - 4 $ |
| $ - 1 $ | $ - 3 $ |
| $ 1 $ | $ - 1 $ |
| $ 2 $ | $ 0 $ |
Vậy ...
Câu 1: \(3^{3x+3}=27\)
\(\Rightarrow3^{3x+3}=3^3\)
\(\Rightarrow3x+3=3\)
\(\Rightarrow x=\dfrac{3-3}{3}=0\)
Vậy \(x=0\)
