Question

Tìm x biết :

a)\(3^{-2}.3^2.27^x=\frac{1}{3}\)

b)\(7^{x+2}+2.7^{x-1}=345\)

c)\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

Answer 1

a.\(3^{-2}.3^2.27^x=\frac{1}{3}\)

\(\Rightarrow3^{-2+2}.\left(3^3\right)^x=\frac{1}{3}\)

\(\Rightarrow3^0.3^{3x}=3^{-1}\)

\(\Rightarrow3^{3x}=3^{-1}\)

=> 3x=-1

=> x=\(-\frac{1}{3}\)

b.\(7^{x+2}+2.7^{x-1}=345\)

\(\Rightarrow7^{x-1}.\left(7^3+2\right)=345\)

\(\Rightarrow7^{x-1}.345=345\)

=> 7^x-1=345 : 345

=> 7^x-1=1

=> 7^x-1=7⁰

=> x-1=0

Vậy x=1.

c.\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)\in\left\{-1;0;1\right\}\)

=> 2x-1=-1               hoặc 2x-1=0                         hoặc 2x-1=1

=> 2x=0                   hoặc 2x=1                            hoặc 2x=2

=> x=0                     hoặc x=\(\frac{1}{2}\)                           hoặc x=1

Vậy \(x\in\left\{0;\frac{1}{2};1\right\}\)