\(3^{x+3}\). \(3^{2x-1}\)\(+\) \(3^{2x}\).\(3^{x+1}\)\(=\)\(324\)
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25 tháng 10 2020
Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)
\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)
\(\Leftrightarrow3^{3x+1}\cdot4=324\)
\(\Leftrightarrow3^{3x+1}=81\)
\(\Leftrightarrow3^{3x+1}=3^4\)
\(\Rightarrow3x+1=4\)
\(\Rightarrow x=1\)
23 tháng 10 2015
32x+1+32x=324
=> 32x.(31+1)=324
=> 32x.4=324
=> 32x=81
=> 32x=34
=> 2x=4
=> x=2
K
10 tháng 11 2018
\(3^{2x}+3^{2x+1}=324\)
\(\Rightarrow3^{2x}+3^{2x}.3=324\)
\(\Rightarrow3^{2x}\left(1+3\right)=324\)
\(\Rightarrow3^{2x}.4=324\)
\(\Rightarrow3^{2x}=324:4=81=3^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2=2\)
10 tháng 11 2018
32x + 32x + 1 = 324
=> 32x + 32x.3 = 324
=> 32x.(1 + 3) = 324
=> 32x.4 = 324
=> 32x = 324 : 4
=> 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 2
DP
0
13 tháng 12 2017
\(3^{2x+1}+9^x=324\)
\(3^{2x}.3+3^{2x}=324\)
\(3^{2x}.\left(3+1\right)=324\)
\(3^{2x}.4=324\)
\(3^{2x}=81\)
\(3^{2x}=3^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
vậy \(x=2\)
DD
13 tháng 12 2017
32x+1 + 9x = 324
9x lớn nhất là 81 trong phép cộng trên
9x = 92 = 81
x = 2
15 tháng 6 2015
9x+1-5.32x=324
=>9x.9-(32)x.5=324
=>9x.9-9x.5=324
=>9x(9-5)=324
=>9x.4=324
=>9x=324:4
=>9x=81
=>9x=92
=>x=2
vậy x=2
15 tháng 7 2019
a) x - 3/97 + x - 2/98 = x - 1/99 + x/100
<=> x + 1/99 + 1 + x + 2/98 + 1 + x + 3/97 + 1 + (x + 4/96 + 1 + x + 5/95 + 1 + x + 10/90 + 1) = 0
<=> x + 100/99 + x + 100/98 + x + 100/97 + (x + 100/96 + x + 100/95 + x + 100/90) = 0
<=> (x + 100)(1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90) = 0
Mà 1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90 khác 0
=> x + 100 = 0
=> x = -100
c) (1/1.2 + 1/2.3 + ... + 1/99.100) - 2x = 1/2
<=> (1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100) - 2x = 1/2
<=> (1 - 1/100) - 2x = 1/2
<=> 99/100 - 2x = 1/2
<=> -2x = 1/2 - 99/100
<=> -2x = -49/100
<=> x = 49/200
=> x = 49/200
15 tháng 7 2019
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
28 tháng 7 2017
a) \(8\left(x-2\right)=3\)
\(\Leftrightarrow x-2=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+2\)
\(\Leftrightarrow x=\dfrac{19}{8}\)
Vậy \(x=\dfrac{19}{8}\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow9^x.9-\left(3^2\right)^x.5=324\)
\(\Rightarrow9^x.9-9^x.5=324\)
\(\Rightarrow9^x\left(9-5\right)=324\)
\(\Rightarrow9^x.4=324\)
\(\Rightarrow9^x=\dfrac{324}{4}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)
\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)
\(\Leftrightarrow3^{3x+1}\cdot4=324\)
\(\Leftrightarrow3^{3x+1}=81=3^4\)
\(\Rightarrow3x+1=4\)
\(\Leftrightarrow x=1\)
\(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(3^{x+3+2x-1}+3^{2x+x+1}=324\)
\(3^{3x+2}+3^{3x+1}=324\)
\(3^{3x+1}\cdot\left(3+1\right)=324\)
\(3^{3x+1}\cdot4=324\)
\(3^{3x+1}=324:4\)
\(3^{3x+1}=81\)
\(3^{3x+1}=3^4\)
\(\Rightarrow3x+1=4\)
\(3x=4-1\)
\(3x=3\)
\(x=3:3\)
\(x=1\)