Question

Bài 1:Tìm x biết:

a,\(9^{x-1}=\dfrac{1}{9}\)                b,\(\dfrac{3^x}{27}=27\)                  c\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-6^x+2=6^7+6^4\)

d,\(2\cdot3^x+3^{x-1}=7\cdot\left(3^2+2\cdot6^2\right)\)

(Giúp mình nhanh nhé,mình đang gấp lắm ạ.)

Answer 1

`a, 9^(x-1)=1/9`

`=> 9^(x-1)=9^(-1)`

`=>x-1=-1`

`=>x=-1+1`

`=>x=0`

Vậy: `x=0`

`b, (3^x)/27 = 27`

`=> (3^x)/(3^3)=3^3`

`=> 3^x = 3^3 . 3^3`

`=> 3^x = 3^6`

`=>x=6`

Vậy: `x=6`

Answer 2

\(a,9^{x-1}=\dfrac{1}{9}\)
\(9^{x-1}=9^{-1}\)
\(x-1=-1\)
\(x=-1+1\)
\(x=0\)
Vậy ....
\(b,\dfrac{3^x}{27}=27\)
\(3^x=27.27\)
\(3^x=729\)
\(3^x=3^6\)
\(x=6\)
Vậy ....
\(d,2.3^x+3^{x-1}=7.\left(3^2+2.6^2\right)\)
\(2.3^x+3^x:3=7.\left(9+2.36\right)\)
\(2.3^x+3^x.\dfrac{1}{3}=7.\left(9+72\right)\)
\(3^x.\left(2+\dfrac{1}{3}\right)=7.81\)
\(3^x.\dfrac{7}{3}=567\)
\(3^x=567:\dfrac{7}{3}\)
\(3^x=243\)
\(3^x=3^5\)
Vậy ....