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

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

\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)

\(\Rightarrow2x-1=0\)hoặc \(2x-1=1\)hoặc \(2x-1=-1\)

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

123+x+2x=2x+345-236

<=>x+2x-2x=345-236-123

<=>x=-14

vậy x=-14

1324+x+x=125-2x-3x

<=>x+x-2x-3x=125-1324

<=>-3x=-1199

<=>x=\(\frac{1999}{3}\)

vậy x=\(\frac{1999}{3}\)

k mk nha mấy bạn

\(123+x+2x=2x+345-236\)

\(\Leftrightarrow2x+345-236-2x-x=123\)

\(\Leftrightarrow109-x=123\)

\(\Leftrightarrow x=-14\)

a) x – 123 = 345                                                         b) x + 254 = 665

         x      = 345 + 123                                                    x          = 665 – 254

         x      =  468                                                                x       = 211

a) x – 345 094 = 123 357

x = 123 357 + 345 094

x = 468 451

b) x : 13 = 125

x = 125 x 13 

x = 1625

a: =>x=123357+345094=468451

b: =>x=125x13=1625

\(a,x-345094=123357\\ \Rightarrow x=123357+345094\\ \Rightarrow x=468451\\ b,x:13=125\\ \Rightarrow x=125\times13\\ \Rightarrow x=1625\)

a)TH1: \(2x-3>0;3x+2>0\)

\(=>2x-3-3x-2=0\\ =>-x-5=0\\ =>-x=5=>x=-5\)

TH2: \(2x-3< 0;3x+2< 0\)

\(=>-2x+3+3x+2=0\\ =>x+5=0\\ =>x=-5\)

Cả 2 TH ra \(x=-5=>x=-5\)

b)TH1 \(\dfrac{1}{2}x>0\)

\(=>\dfrac{1}{2}x=3-2x\\ =>3-2x-\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x-\dfrac{1}{2}x=3\\ =>\dfrac{3}{2}x=3\\ =>x=2\)

TH2 \(\dfrac{1}{2}x< 0\)

\(=>-\dfrac{1}{2}x=3-2x\\ =>3-2x+\dfrac{1}{2}x=0\\ =>\dfrac{4}{2}x+\dfrac{1}{2}x=3\\ =>\dfrac{5}{2}x=3\\ =>x=\dfrac{6}{5}\)

\(=>x=2;\dfrac{6}{5}\)

:V lập 2 ý là làm đc á em 

xét 2 trường hợp là đc mà e :v