a) \(20\cdot2^x+1=10\cdot4^2+1\)
\(\Leftrightarrow2\cdot10\cdot2^x=10\cdot4^2\)
\(\Leftrightarrow10\cdot2^{x+1}=10\cdot2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b) \(\left(4-\frac{x}{2}\right)^3-1=2\cdot\left(2^3-\frac{5}{2^0}\right)+1\)
\(\Leftrightarrow\left(4-\frac{x}{2}\right)^3=2\cdot3+1+1\)
\(\Leftrightarrow\left(4-\frac{x}{2}\right)^3=8=2^3\)
\(\Rightarrow4-\frac{x}{2}=2\)
\(\Leftrightarrow\frac{x}{2}=2\)
\(\Rightarrow x=4\)
c) Ta xét 2 TH sau:
Nếu n lẻ
=> n + 2017 chẵn
=> n(n+2017) chẵn (1)
Nếu n chẵn
=> n(n+2017) chẵn (2)
Từ (1) và (2) => n(n+2017) luôn chẵn với mọi STN n
=> đpcm
d) Ta có:
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Mà \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
Vậy 3200 > 2300
a)\(20.2^x +1= 40^2+1 \)
\(20.2^x +1=1601\)
\(40^x=1601-1 \)
\(40^x=1600\)
x=2
Cho mình xin lỗi!!!!!!!!!!!!!!!!!
a,\(20.2^x+1=10.4^2+1\)
\(20.2^x=10.4^2\)
\(10.2.2^x=10.2^4\)
\(10.2^{x+1}=10.2^4\)
\(2^{x+1}=2^4\)
\(x+1=4\)
\(x=3\)