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

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

\(\Rightarrow9x+3=117\)

\(\Rightarrow9x=117-3\)

\(\Rightarrow9x=114\)

\(\Rightarrow x=114:9\)

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

Vậy \(x=\frac{38}{3}\)

P/s : Đúng nha 

~ Ủng hộ nhé 

=> 6x + 3= 117

6x = 117-3

6x= 114

x=19

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) \(3+4\left(x-10\right)=3^2+6\)

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

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

\(4x=52\)

\(x=13\)

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

9x+3=117

9x=117-3

9x=114

x=114/9

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

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

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

\(\text{Vậy x=}\frac{114}{9}\)

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

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

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

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

\(\Rightarrow x.9=114\)

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

Vậy \(x=\dfrac{38}{3}\)

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

=>9x+3=117

=>9x=117-3=114

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

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=\(\frac{35}{3}\)

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

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

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: \(\dfrac{-3x-4y}{-15-24}=\dfrac{-117}{-39}=3\)

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

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

a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)

\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)

\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)

c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)

\(=\dfrac{-x+7x-4}{3x-2}\)

\(=\dfrac{6x-4}{3x-2}=2\)