a) \(2^x\cdot2^2-2^x=96\)
\(2^x\cdot\left(4-1\right)=96\)
\(2^x=32=2^5\)
=> x = 5
b) \(\left(2x+1\right)^3=7^3\)
=> 2x +1 = 7
=> 2x = 6
=> x = 3
Vậy,..........
a. 2x+2 - 2x = 96
2x . 22 - 2x = 96
2x. (22 - 1) = 96
2x = 96 : 3
2x = 32
2x = 25
x = 5
b. (2x + 1)3 = 343
(2x + 1)3 = 73
2x + 1 = 7
2x = 7 - 1
x = 6 : 2
x = 3
a) \(2^{x+2}\)−\(2^x\)=96
\(\Rightarrow\)\(2^x\)·(4−1)=96
\(\Rightarrow2^x\)=32=\(2^5\)
\(\Rightarrow\)\(x=5\)
b) (2x+1)\(^3\)=343
\(\Rightarrow2x+1=7\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
