3^x.3^3=243
3^x + 3 ^x+1 + 3^x+2= 243
3x + 3x+1 + 3x+2 = 243
3x . 1 +3x . 31 + 3x .32 = 243
3x .(1 + 3 + 32) = 243
3x .(1 +3 + 9) = 243
3x . 13 = 243
3x = 243 : 13
Vì 243 : 13 không thuộc N => x thuộc rỗng
3^x-4.(3^2)^3=243
đề bài có phải như này không em ?
\(3^{x-4.\left(3^2\right)^3}=243\)
< = > \(3^{x-4.\left(3^2\right)^3}=3^5\)
< = > x - 4 . ( 32 )3 = 5
< = > x - 4 . 729 = 5
< = > x - 2916 = 5
< = > x = 2921
\(...\Rightarrow3^{x-4}.3^6=2023\)
\(\Rightarrow3^{x-4+6}=3^5\)
\(\Rightarrow3^{x+2}=3^5\Rightarrow x+2=5\Rightarrow x=3\)
a) 32x-1= 243 b) (3x)2 :33= 1/243
c) 23x+2= 4x+5 d) 3x+1= 9x
a) \(3^{2x-1}=243\)
\(\Leftrightarrow3^{2x-1}=3^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(3^x\right)^2:3^3=\dfrac{1}{243}\)
\(\Leftrightarrow3^{2x}:3^3=\dfrac{1}{3^5}\)
\(\Leftrightarrow3^{2x}:3^3=3^{-5}\)
\(\Leftrightarrow3^{2x-3}=3^{-5}\)
\(\Leftrightarrow2x-3=-5\)
\(\Leftrightarrow2x=-5+3\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{2}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
c) \(2^{3x+2}=4^{x+5}\)
\(\Leftrightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Leftrightarrow2^{3x+2}=2^{2\left(x+5\right)}\)
\(\Leftrightarrow3x+2=2\left(x+5\right)\)
\(\Leftrightarrow3x+2=2x+10\)
\(\Leftrightarrow3x-2x=10-2\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
d) \(3^{x+1}=9^x\)
\(\Leftrightarrow3^{x+1}=\left(3^2\right)^x\)
\(\Leftrightarrow3^{x+1}=3^{2x}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow2x-x=1\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\((\dfrac{1}{3})^x+(\dfrac{1}{3})^{x+3}=\dfrac{28}{243}\)
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+3}=\dfrac{28}{243}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\left(\dfrac{1}{3}\right)^3=\dfrac{28}{243}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\dfrac{1}{27}=\dfrac{28}{243}\\ \Rightarrow\dfrac{28}{27}\left(\dfrac{1}{3}\right)^x=\dfrac{28}{243}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x=\dfrac{28}{243}:\dfrac{28}{27}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x=\dfrac{1}{9}\\ \Rightarrow\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^2\\ \Rightarrow x=2\)
(1/3)x +(1/3)x+3=28/243 (1/3)x +(1/3)x *(1/3)3=28/243 (1/3)x *(1+1/33)=28/243 (1/3)x *28/27=28/243 (1/3)x=1/9 (1/3)x=(1/3)2 vậy x =2
\((\dfrac{1}{3})^x+(\dfrac{1}{3})^{x+3}=\dfrac{28}{243}\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^x\cdot\dfrac{28}{27}=\dfrac{28}{243}\)
hay x=2
(3^x)^2 : 3^3=1/243
tìm x
(3^x)^2=1/243.3^3
(3^x)^2=1/9
3x=1/3 hoặc 3x=-1/3
Suy ra x=1/9 hoặc x=-1/9
3 mũ x+1 +3 mũ 2=243 tìm x
\(\Leftrightarrow3^x\cdot3=243\)
hay x=4
3^x . 3 = 243 tìm x nhé
3x.3=243
<=>3x.31=243
Mà: 35=243
=> 3x.31=35
=>3x= 35-1
=>3x=34
=> x=4
Ta có : 3^x . 3 =243
=>3^x = 243 : 3 = 81
=>3^x = 3^4
=>x = 4
3x x 3 = 243
3x x 3 = 243
Toán lớp 6
3x.3=243
3x=243:3
3x=81
3x=34
x = 4