2^{x + 3} - 3 . 2^{x + 1} = 32

2^x . 2³ - 3 . 2^x . 2 = 32

2^x ( 8 - 3 . 2 ) = 32

2^x . 2 = 32

2^x = 32 : 2 = 16 = 2⁴

Vậy x = 4 

x = { 0 ; 2 ; 4 ; 6 ; 8 }

2+3🤔

1+2+3

a)9 và 7

b)1

chi tiết hơn được không

`5^{3}.5^{x-2}=125^{2}`

`=>5^{3+x-2}=(5^{3})^{2}`

`=>5^{x+1}=5^{3.2}=5^{6}`

`=>x+1=6`

`=>x=5`

\(5^3.5^{x-2}=125^2\)

\(\Leftrightarrow125.5^{x-2}=125^2\)

\(\Leftrightarrow5^{x-2}=125\)

\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)

`140:(x-8)=7`

`=>x-8=140:7`

`=>x-8=20`

`=>x=20+8`

`=>x=28`

``

`720:(41-x)=2^{3}.5`

`=>720:(41-x)=8.5=40`

`=>41-x=720:40=18`

`=>x=41-18`

`=>x=23`

+) \(140:\left(x-8\right)=7\)

\(\Leftrightarrow x-8=20\)

\(\Leftrightarrow x=28\)

+) \(720:\left(41-x\right)=2^3.5\)

\(\Leftrightarrow41-x=18\)

\(\Leftrightarrow x=23\)

( x - 31 ) + 27 = 94

x - 31 = 94 - 27

x - 31 = 67

x = 67 + 31

x = 98

\(\left(x-31\right)+27=94\)

\(\Leftrightarrow x-31=67\)

\(\Leftrightarrow x=98\)

\(2019^0+\left[1300-\left(4^2-2.3\right)^2\right]:40\)

\(=1+\left[1300-\left(16-6\right)^2\right]:40\)

\(=1+\left(1300-100\right):40\)

\(=1+1200:40\)

\(=1+30=31\)

2019⁰ + [ 1300 - ( 4² - 2 .3 )² ] : 40

= 1 + [ 1300 - ( 16 - 6 )² ] : 40

= 1 + [ 1300 - 10² ] : 40

= 1 + [ 1300 - 100 ] : 40

= 1 + 1200 : 40

= 1 + 30 = 31

Ư( 28 ) = { 1 ; 2 ; 4 ; 7 ; 14 ; 28 }

Ư( 21 ) = { 1 ; 3 ; 7 ; 21 }

Vậy x ϵ { 1 ; 7 }