\(\left(x+2\right)-34=27-\left(2x+1\right)\)

\(\Rightarrow x+2-34=27-2x-1\)

\(\Rightarrow x-32=26-2x\)

\(\Rightarrow x+2x=26+32\)

\(\Rightarrow3x=58\)

\(\Rightarrow x=58\div3\)

\(\Rightarrow x=19,\left(3\right)\)

X +2 -34 =27 -2X -1

X -32 = 26 -2X

X -32 -26 +2X =0

3X-58=0

a ) x − 2 5 : 2 3 = 3 4 x − 2 5 = 3 4 . 2 3 x − 2 5 = 1 2 x = 9 10

b ) x + − 3 4 − 1 2 = 2 3 . 3 5    x + − 3 4 − 1 2 = 2 5 x + − 3 4 = 9 10 x = 33 20

c ) x − 2 7 = − 9 14 . − 4 3 x − 2 7 = 6 7 +)  x − 2 7 = 6 7 x = 8 7 +)  x − 2 7 = − 6 7    x = − 4 7

1: Tìm x

a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)

\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)

\(\Leftrightarrow3^x=3^{12}\cdot3^3\)

\(\Leftrightarrow3^x=3^{15}\)

hay x=15

Vậy: x=15

b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)

\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)

Vậy: x=2

b)-3.!x-3!=-27

!x-1!=-27/-3

!x-1!=9

x-1=+-9

TH1 TH2

x-1=9 x-1=-9

x=9+1 x=-9+1

x=10 x=-8

a, = 34.16 - 34.44 + 44.34 - 44.16

= 16.(34 - 44 ) + 44 .( 34 - 34 )

= 16.(-10) + 44.0

= -160

b, ! x-1 ! = -27 : (-3)

! x-1 ! = 9

TH1 : x-1=9 => x= 10

TH2 : x-1 = -9 => x = -8