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

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

\(\Rightarrow9x+3=117\)

\(\Rightarrow9x=117-3\)

\(\Rightarrow9x=114\)

\(\Rightarrow x=114:9\)

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

Vậy \(x=\frac{38}{3}\)

P/s : Đúng nha 

~ Ủng hộ nhé 

=> 6x + 3= 117

6x = 117-3

6x= 114

x=19

\(\left|-2\right|\cdot3^x+3^{x+2}=99\)

\(\Rightarrow2\cdot3^x+3^x\cdot3^2=99\)

\(\Rightarrow3^x\cdot\left(2+3^2\right)=99\)

\(\Rightarrow3^x\cdot11=99\)

\(\Rightarrow3^x=99:11\)

\(\Rightarrow3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

\(2\cdot3^x-405=3^{x-1}\\\Rightarrow2\cdot3^x-3^{x-1}=405\\\Rightarrow2\cdot3^x-3^x\cdot3^{-1}=405\\\Rightarrow3^x\cdot(2-3^{-1})=405\\\Rightarrow3^x\cdot(2-\frac13)=405\\\Rightarrow3^x\cdot\frac53=405\\\Rightarrow3^x=405:\frac53\\\Rightarrow3^x=243\\\Rightarrow3^x=3^5\\\Rightarrow x=5\)

`-3x=2y `

`=> x/2 = -y/3 `

AD t/c của dãy tỉ số bằng nhau ta có

`x/2 =-y/3 = (x-y)/(2+3) = 6/5`

`=>{(x=2*6/5 = 12/5),(y=-3*6/5 =-18/5):}`

a) `6/x =-3/2`

`=>x =6 :(-3/2) = 6*(-2/3)=-4`

`b)`\(-3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)

Áp dụng t/c của DTSBN , ta đc :

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2+3}=\dfrac{6}{5}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{6}{5}\\\dfrac{y}{-3}=\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\y=-\dfrac{18}{5}\end{matrix}\right. \)

`a)`

`6/x=-3/2`

`x=6:(-3/2)`

`x=6*(-2/3)`

`x=-4`

hk hừng đẳng thưc chưa

chưa bn

Làm sai