a) x20 = x
=> x20 - x = 0
=> x(x19 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^{19}=1^{19}\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x = 0 hoặc x = 1
b) 3x + 2 - 5.3x = 36
=> 3x . 32 - 5.3x = 36
=> 3x.9 - 5.3x = 36
=> 3x.(9 - 5) = 36
=> 3x.4 = 36
=> 3x = 9
=> 3x = 32
=> x = 2
Vậy x = 2
\(a,\text{ }x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x^{19}=0+1=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }1\right\}\)
\(a,\text{ }x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x^{19}=0+1=1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\text{Vậy }x\in\left\{0\text{ ; }1\right\}\)
\(b,\text{ }3^{x+2}-5\cdot3^x=36\)
\(3^x\cdot3^2-5\cdot3^x=36\)
\(3^x\cdot9-5\cdot3^x=36\)
\(3^x\left(9-5\right)=36\)
\(3^x\cdot4=36\)
\(3^x=36\text{ : }4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow\text{ }x=2\)
\(b,\text{ }3^{x+2}-5\cdot3^x=36\)
\(3^x\left(3^2-5\right)=36\)
\(3^x\left(9-5\right)=36\)
\(3^x\cdot4=36\)
\(3^x=36\text{ : }4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow\text{ }x=2\)
x20=x
x20-x=0
x(x19-1)=0
Trường hợp 1:x=0
Trường hợp 2:x19-1=0
x19=1
x=1
Vậy x=1 hoặc x=0
b, 3x+2-5.3x=36
3x(32-5)=36
3x.4=36
3x=9
3x=32
x=2
Vậy x=2
