\(34.3^x.3^{x+1}=243\)
tim x thuoc n: 34.3^n :9=37
tim x thuoc n: 9<3^n<27
Tim chu so tan cung cua 360
hãy so sánh a=1+3+3^2+3^3+3^4+35+36 và B=37-1
a/ 34 . 3n : 9 = 34 => 34 . 3n = 34 x 9 => 34 . 3n = 306 => 3n = 306 : 34 => 3n = 9 => n = 2
b/ 9 < 3n < 27 => 32 < 3n < 33 => 2 < n < 3
Mà: n thuộc N => n không tồn tại
c/ Chữ số tận cùng của 360 là 0
d/ Ta có: A = 1 + 3 + 32 + 33 + 34 + 35 + 36
=> 3A = 3 + 32 + 33 + 34 + 35 + 36 + 37
=> 3A - A = 2A = (3 + 32 + 33 + 34 + 35 + 36 + 37) - (1 + 3 + 32 + 33 + 34 + 35 + 36 ) = 3 + 32 + 33 + 34 + 35 + 36 + 37 - 1 - 3 - 32 - 33 - 34 - 35 - 36
=> 2A = 37 - 1 => A = (37 - 1) : 2 < 37 - 1 = B
=> A < B
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
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
243<3^x-1<2430
(x-1 nhé)
Ta có: 3^5<3^x-1<3^5.10
Vậy x = 7
(1/3)^x.(1/3)^x-2=10/243
3 mũ x+1 +3 mũ 2=243 tìm x
\(\Leftrightarrow3^x\cdot3=243\)
hay x=4
Tìm x
3 mũ x nhân 3 mũ x+1 =243
3x . 3x+1 = 243
<=> 3x+x+1 = 35
<=> 2x+1 = 5
<=> x = 2
3x x 3x+1 = 243
3x x 3x+1 = 35
=> X x X+1=5
=> x2 = 5+1
=> x2 = 6
=> x=6
Vậy x = 6
# Chúc bạn học tốt #
3x.3x+1=243
3x.3x+31=243
3x.31=243
3x=243:3
3x=81
3x=34
x=4