Bài 1 :
a, \(2^x+2^{x+1}=24\)
\(\Rightarrow2^x.1+2^x.2=24\)
\(\Rightarrow2^x\left(1+2\right)=24\)
\(\Rightarrow2^x=24\div3\)
\(\Rightarrow2^x=8=2^3\)
Vậy : x = 3
b, \(x^2-x=0\)
\(\Rightarrow x.x-x.1=0\)
\(\Rightarrow x\left(x-1\right)=0\)
Để : \(x\left(x-1\right)=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy x = 1
Bài 2 :
a, \(Q=3+3^3+3^5+...+3^{101}\)
\(\Rightarrow9Q=3^3+3^5+3^7+...+3^{103}\)
\(\Rightarrow9Q-Q=\left(3^3+3^5+3^7+...+3^{103}\right)-\left(3+3^3+3^5+...+3^{101}\right)\)
\(\Rightarrow8Q=3^{103}-3\)
\(\Rightarrow Q=\frac{3^{103}-3}{8}\)
b, \(Q=3+3^3+3^5+...+3^{101}\)
\(\Rightarrow Q=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{97}+3^{99}+3^{101}\right)\)
\(\Rightarrow Q=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{96}\left(3+3^3+3^5\right)\)
\(\Rightarrow Q=1.273+3^6.273+...+3^{96}.273\)
\(\Rightarrow Q=\left(1+3^6+...+3^{96}\right)273\)
Vì : \(1+3^6+...+3^{96}\in N\) ; \(273=3.91\Rightarrow Q⋮91\)
Vậy ...
