\(3x+3x+1+3x+2=117\)

\(\Rightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)

\(\Rightarrow9x+3=117\)

\(\Rightarrow9x=117-3\)

\(\Rightarrow9x=114\)

\(\Rightarrow x=114:9\)

\(\Rightarrow x=\frac{38}{3}\)

Vậy \(x=\frac{38}{3}\)

P/s : Đúng nha 

~ Ủng hộ nhé 

=> 6x + 3= 117

6x = 117-3

6x= 114

x=19

=>9x +3 =117

9x=114

x=38/3

a)

\(3^x+3^{x+1}+3^{x+2}=117\\ \Leftrightarrow3^x+3.3^x+9.3^x=117\\ 13.3^x=117\\ \Leftrightarrow3^x=9\\ \Leftrightarrow3^x=3^2\\ \Leftrightarrow x=2\)

b)

 \(3+4\left(x-10\right)=3^2+6\\ \Leftrightarrow3+4\left(x-10\right)=15\\ \Leftrightarrow4\left(x-10\right)=12\\ \Leftrightarrow x-10=3\\ \Leftrightarrow x=13\)

a) \(3^x+3^{x+1}+3^{x+2}=117\)

\(3^x+3^x.3+3^x.3^2=117\)

\(3^x.\left(1+3+3^2\right)=117\)

\(3^x.13=117\)

\(3^x=9\)

\(x=2\)

b) \(3+4\left(x-10\right)=3^2+6\)

\(3+4x-40=9+6\)

\(4x=15+40-3\)

\(4x=52\)

\(x=13\)

x = 0

y = 29\(\dfrac{1}{4}\)

\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{-3x}{-15}=\dfrac{4y}{24}\)

Áp dụng t/c của DS bằng nhau, ta có: \(\dfrac{-3x-4y}{-15-24}=\dfrac{-117}{-39}=3\)

\(\dfrac{-3x}{-15}=3\Rightarrow x=15\)

\(\dfrac{4y}{24}=3\Rightarrow y=18\)

3x+3x+1+3x+2=177

9x+3=117

9x=117-3

9x=114

x=114/9

\(3x+3x+1+3x+2=117\)

\(\Leftrightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)

\(\Leftrightarrow9x+3=117\)\(\Rightarrow9x=114\Rightarrow x=\frac{114}{9}\)

\(\text{Vậy x=}\frac{114}{9}\)

3x+3x+1+3x+2=117

=>3x.3 +(1+2) = 117

=>9x +3 = 117

=> 9x = 114

=> x = \(\frac{38}{3}\)

\(3^x+3^{x+1}+3^{x+2}=117\)

\(3^x+3^x.3+3^x.3^2=117\)

\(3^x\left(1+3+3^2\right)=117\)

\(3^x.13=117\)

\(3^x=9\)

\(\Rightarrow x=2\)

`3^{x}+3^{x+1}+3^{x+2}=117`

`3^{x}.(1+3+3^{2})=117`

`3^{x}.13=117`

`3^{x}=117:13=9`

`3^{x}=3^{2}`

`x=2`

\(3^x+3^{x+1}+3^{x+2}=117\)

\(3^x\left(1+3+3^2\right)=117\)

\(3^x\times13=117\)

\(3^x=9\)

\(3^x=3^2\)

\(x=2\)

Bài 1:

Thay \(x=\frac{4}{3};y=-1\)vào biểu thức A, ta được:

\(A=\frac{4}{3}\cdot\left[3\cdot\frac{4}{3}-\left(-1\right)\right]-\left(3\cdot\frac{4}{3}+1\right)\left(-1\right)\)

\(A=\frac{20}{3}+5=\frac{35}{3}\)

Vậy khi \(x=\frac{4}{3};y=-1\)thì A=\(\frac{35}{3}\)

\(B=3\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{8}{39}\)

\(B=\frac{352}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot\frac{713}{119}-\frac{8}{39}=-\frac{412}{1071}\)

\(3x+3x+1+3x+2=117\)

\(\Rightarrow3x+3x+3x=117-1-2\)

\(\Rightarrow3x+3x+3x=114\)

\(\Rightarrow x.\left(3+3+3\right)=114\)

\(\Rightarrow x.9=114\)

\(\Rightarrow x=\dfrac{38}{3}\)

Vậy \(x=\dfrac{38}{3}\)

=> 3x+3x+3x+1+2=117

=>9x+3=117

=>9x=117-3=114

=> x=\(\dfrac{114}{9}\)

\(3x+3x+1+3x+2=117\)

\(\Leftrightarrow9x=114\)

\(\Leftrightarrow x=\dfrac{38}{3}\)

đặt 3^x ra ngoài bạn nhé bàn phím mik hỏng rồi ;-;

$\Rightarrow 3^x(1+3+3^2+3^3)=1080$

$\Rightarrow 3^x.40=1080$

$\Rightarrow 3^x=27=3^3$

$\Rightarrow x=3$

\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=1080\)

\(\Rightarrow3^x+3^x.3^1+3^x.3^2+3^x.3^3=1080\)

\(3^x.\left(1+3+9+27\right)=1080\)

\(3^x.40=1080\)

\(3^x=27\)

\(3^x=3^3\)

Vậy: x = 3

Ta sẽ đưa các tích về 1 dãy tỉ số

\(3x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{3}\Leftrightarrow\frac{x}{15}=\frac{y}{9},7y=9z\Leftrightarrow\frac{y}{9}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{9}=\frac{z}{7},x-y+z=117\left(gt\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau cho dãy tỉ số trên ta được

\(\frac{x}{15}=\frac{y}{9}=\frac{z}{7}=\frac{x-y+z}{15-9+7}=\frac{117}{13}=9\Rightarrow x=15.9=135,y=9.9=81,z=7.9=63\)

Vậy \(x=135,y=81,z=63\)

Ta có: \(3x=5y=\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x}{15}=\frac{y}{9}\)
             \(7y=9z=\frac{y}{9}=\frac{z}{7}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{9}=\frac{z}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{15}=\frac{y}{9}=\frac{z}{7}=\frac{x-y+z}{15-9+7}=\frac{117}{13}=9\)

\(\Rightarrow\frac{x}{15}=9\Rightarrow x=9\cdot15=135\)

       \(\frac{y}{9}=9\Rightarrow y=9\cdot9=81\)

       \(\frac{z}{7}=9\Rightarrow z=9\cdot7=63\)

Vậy x=135, y=81 và z=63

a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow9x^2-3x-9x^2+1=0\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)

b: Ta có: \(x^2-5x+25-5x=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

hay x=5