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\(3^{x+2}+4\cdot3^{x+1}=7\cdot3^6\\ 3^x\cdot3^2+4\cdot3^x\cdot3=5103\\ 3^x\left(9+12\right)=5103\\ 3^x\cdot21=5103\\ 3^x=243\\ 3^x=3^5\\ x=5\)
\(4.3^{x+2}-3^{x-1}=963\)
\(\Rightarrow4.3^{x+2}-3^{x+2-3}=963\)
\(\Rightarrow4.3^{x+2}-\dfrac{1}{3^3}3^{x+2}=963\)
\(\Rightarrow4.3^{x+2}-\dfrac{1}{27}3^{x+2}=963\left(1\right)\)
Đặt \(t=3^{x+2}>0\)
\(\left(1\right)\Rightarrow4t-\dfrac{1}{27}t=963\)
\(\Rightarrow t\left(4-\dfrac{1}{27}\right)=963\)
\(\Rightarrow\dfrac{107}{27}.t=9.107\)
\(\Rightarrow t=3^2\Rightarrow t=3^2:\dfrac{1}{3^3}=3^2.\dfrac{3^3}{1}=3^5\)
\(\Rightarrow3^{x+2}=3^5\Rightarrow x+2=5\Rightarrow x=3\)
4\(\cdot\)3x+2-3x-1=963
=>4\(\cdot\)3xx32-3x:3=963
=>3x\(\cdot\)36-3x\(\cdot\)\(\dfrac{1}{3}\)=963
=>3x\(\cdot\)(36-\(\dfrac{1}{3}\))=963
=>3x\(\cdot\)\(\dfrac{107}{3}\)=963
=>3x=963:\(\dfrac{107}{3}\)=27=33
=>x=3
Ta có: \(4.3^{x+2}-2.3^{x+1}=810\)
\(4.3^{x+1}.3-2.3^{x+1}=810\)
\(2.3^{x+1}\left(2.3-1\right)=810\)
\(2.3^{x+1}.5=810\)
\(3^{x+1}=81=3^4\)
x+1=4
x=3
Vì 963 có 9+6+3 = 18 chia hết cho 9 => 963 chia hết cho 9
2493 có 2+4+9+3 = 18 chia hết cho 9 => 24933 chia hết cho 9
351 có 3+5+1 =9 chia hết cho 9
+Để A chia hết cho => x chia hết cho 9
+Để A không hia hết cho 9 => x không chia hết cho 9
\(\left(4.3x-4\right)-2=18\)
\(\Rightarrow12x-4-2=18\)
\(\Rightarrow12x=24\Leftrightarrow x=2\)
a ) 9 . 3x = 81
3x = 9
x = 3
b ) 2x : 4 = 1
2x = 4
x = 2
c ) 2x - 64 = 2
2x = 66
x = 33
d ) 2x = 16
x = 8
e ) 3 ^ 2 . 3 ^ 4 . 3x = 3 ^ 10
3 ^ ( 2 + 4 + x ) = 3 ^ 10
=> 2 + 4 + x = 10
x = 4
f ) 2x + 4 . 2x = 5 . 2 ^ 5
2x + 8 x = 160
10x = 160
x = 16
\(4\cdot3^{x+2}-3^{x-1}=963\\ \Leftrightarrow36\cdot3^x-\dfrac{1}{3}\cdot3^x=963\\ \Leftrightarrow3^x=27\\ \Leftrightarrow x=3\)
\(4.3^{x+2}-3^{x-1}=963\\ \Leftrightarrow4.3^2.3^x-\dfrac{3^x}{3}=963\\ \Leftrightarrow3^x.\left(36-\dfrac{1}{3}\right)=963\\ \Leftrightarrow3^x.\dfrac{107}{3}=963\\ \Leftrightarrow3^x=27\\ \Leftrightarrow3^x=3^3\\ \Leftrightarrow x=3\)
Vậy x= 3