a, => 3^x.(6+4.3) = 100-72
=> 3^x.18 = 18
=> 3^x = 18:18 = 1
=> 3^x = 3^0
=> x=0
b, => 5^x.(5+4.5) = -12+175-38
=> 5^x.25 = 125
=> 5^x = 125:25 = 5
=> 5^x = 5^1
=> x=1
Tk mk nha
a) \(6\cdot3^x+4\cdot3^{x+1}=100+\left(-72\right)\)
\(\Leftrightarrow6\cdot3^x+4\cdot3^x\cdot3=100-72\)
\(\Leftrightarrow6\cdot3^x+12\cdot3^x=28\)
\(\Leftrightarrow18\cdot3^x=28\Leftrightarrow3^x=\frac{14}{9}\)
Phần a) bn xem lại đề bài nhé!!
b) \(5\cdot5^x+4\cdot5^{x+1}=\left(-12\right)+175+\left(-38\right)\)
\(\Leftrightarrow5^{x+1}+4\cdot5^{x+1}=175-12-38\)
\(\Leftrightarrow5\cdot5^{x+1}=125\Leftrightarrow5^{x+1}=25\)
\(\Leftrightarrow5^{x+1}=5^2\Leftrightarrow x+1=2\Leftrightarrow x=1\)
Vậy x=1
