a/ \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(\Leftrightarrow2^x\left(\dfrac{1}{2}+4\right)=9.32\)
\(\Leftrightarrow2^x.\dfrac{9}{2}=288\)
\(\Leftrightarrow2^x=64\)
\(\Leftrightarrow2^x=2^6\)
\(\Leftrightarrow x=6\)
Vậy ....
b/ \(3^{x+1}-3^{x-2}-3^x=153\)
\(\Leftrightarrow3^x.3^1-3^x:3^2-3^x=153\)
\(\Leftrightarrow3^x.3-3^x.\dfrac{1}{9}-3^x=153\)
\(\Leftrightarrow3^x\left(3-\dfrac{1}{9}-1\right)=153\)
\(\Leftrightarrow3^x.\dfrac{17}{9}=153\)
\(\Leftrightarrow3^x=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Leftrightarrow x=4\)
Vậy ..
a) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
⇒ \(\left(\dfrac{1}{2}+4\right).2^{2x}=9.32\)
⇒ \(\dfrac{9}{2}.2^{2x}=288\)
⇒ \(2^{2x}=288:\dfrac{9}{2}\)
⇒ \(2^{2x}=64\)
Ta có : \(2^6=64\)
Mà \(2^{2x}=64\)
⇒ x = 6 : 2
x = 3
