Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)

\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)

\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)

\(\Leftrightarrow3^{3x+1}\cdot4=324\)

\(\Leftrightarrow3^{3x+1}=81\)

\(\Leftrightarrow3^{3x+1}=3^4\)

\(\Rightarrow3x+1=4\)

\(\Rightarrow x=1\)

3^2x+1+3^2x=324

=> 3^2x.(3¹+1)=324

=> 3^2x.4=324

=> 3^2x=81

=> 3^2x=3⁴

=> 2x=4

=> x=2

\(3^{2x}+3^{2x+1}=324\)

\(\Rightarrow3^{2x}+3^{2x}.3=324\)

\(\Rightarrow3^{2x}\left(1+3\right)=324\)

\(\Rightarrow3^{2x}.4=324\)

\(\Rightarrow3^{2x}=324:4=81=3^4\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2=2\)

3^2x + 3^{2x + 1} = 324

=> 3^2x + 3^2x.3 = 324

=> 3^2x.(1 + 3) = 324

=> 3^2x.4 = 324

=> 3^2x = 324 : 4

=> 3^2x = 81

=> 3^2x = 3⁴

=> 2x = 4

=> x = 2

3^2x+3^2x+1=324

<=>3^2x+3^2x.3=324

<=>3^2x.(3+1)=324

<=>3^2x.4=324

<=>3^2x=81

<=>3^2x=3⁴

<=>2x=4

<=>x=2

vậy x=2

k mik nhé

hok tốt

\(3^{2x+1}+9^x=324\)

\(3^{2x}.3+3^{2x}=324\)

\(3^{2x}.\left(3+1\right)=324\)

\(3^{2x}.4=324\)

\(3^{2x}=81\)

\(3^{2x}=3^4\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

vậy \(x=2\)

3^2x+1 + 9^x = 324

9^x lớn nhất là 81 trong phép cộng trên

9^x = 9² = 81

x = 2

x=2 nhé

k nhé ai k mình thì mình k lại

9^x+1-5.3^2x=324

=>9^x.9-(3²)^x.5=324

=>9^x.9-9^x.5=324

=>9^x(9-5)=324

=>9^x.4=324

=>9^x=324:4

=>9^x=81

=>9^x=9²

=>x=2

vậy x=2

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

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

nhìu quá vậy từng câu thôi

vâng cảm ơn nhờ bạn giải giúp mình với :(

bình tỉnh lại coi

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 \(x=2\)

a, \(8\left(x-2\right)=3\)

\(\Rightarrow x-2=\dfrac{3}{8}\Rightarrow x=\dfrac{19}{8}\)

b, \(9^{x+1}-5.3^{2x}=324\)

\(\Rightarrow9^x.9-5.9^x=324\)

\(\Rightarrow4.9^x=324\Rightarrow9^x=81=9^2\)

Vì \(9\ne\pm1;9\ne0\) nên \(x=2\)

Chúc bạn học tốt!!!

a) 8.(x-2)=3.

=> (x-2)= 3:8 = 0.375.

x = 0.375 + 2 = 2.375.

b) 9x+1 - 5.32x = 324.

<=> 9.9^x - 5.3^2x=324.

<=> 9.3^2x - 5.3^2x=324.

<=> 3^2x.(9-5)= 3^2x.4= 324.

<=> 3^2x=324:4=81=3⁴.

<=>2x = 4.

<=> x= 2.