\(65-4^{x+2}=2014^0\)
\(4^{x+2}=65-1\)
\(4^{x+2}=64\)
\(4^{x+2}=4^3\)
\(\Rightarrow x+2=3\)
\(x=1\)
Vậy \(x=1\)
Tham khảo nhé~
65 - 4^ x+2 = 2014^0
=>65 - 4^x+2=1
=> 4^x+2=65-1
=> 4^x+2=64
=> 4^x+2=4^3
=> 4^x =4^3-2
=> 4^x =4^1
=> x =1
có :
65 - 4x+2 = 20140
65 - 4x+2=1
4x+2 = 64
4x + 2 = 43
x+2 = 3
x = 1
vậy x = 1
\(65-4^{x+2}=2014^0\)
\(\Leftrightarrow65-4^{x+2}=1\)
\(\Leftrightarrow4^{x+2}=65-1\)
\(\Leftrightarrow4^{x+2}=64\)
\(\Leftrightarrow4^{x+2}=4^3\)
\(\Leftrightarrow x+2=3\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
~~~~~~~~Hok tốt~~~~~~~~
\(65-4^{x+2}=2014^0\)
\(\Rightarrow65-4^{x+2}=1\)
\(\Rightarrow4^{x+2}=65-1\)
\(\Rightarrow4^{x+2}=64\)
\(\Rightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2\)
\(\Rightarrow x=1\)
Vậy x = 1
_Chúc bạn học tốt_