Question

a, 6 . 3^x + 4 . 3^x+1 = 100 + (-72)

b, 5 . 5^x + 4 . 5^x+1 = (-12) + 175 + (-38)

Các bạn giúp mình nhé.

Answer 1

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

Answer 2

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