12.2+12.8=12.(2+8)=12.10=120

12 x 2 + 12 x 8 

= 12 x  ( 2 + 8 )

= 12 x 10

= 120

k mifnh nha bajn

\(12\cdot2+12\cdot8\)

\(=12\cdot\left(2+8\right)\)

\(=12\cdot10\)

\(=120\)

12.2+12.5

=12.(2+5)

=12.7

=84

12.2+12.5=12.(2+5)=12.7=84 tick nha

12.2 + 12. 5    = 12 . ( 2 + 5 )

                      = 12 . 10

                       = 120

-12+36-2=12

12=12

ban coi ki laoi de coi chung de sai do cho 4x

mình giải đc rồi cảm ơn bạn nha

(2^4.<20+12>-48.2^2):8^2

=(2^4.32-192):8^2

=320:8^2

=5

(12x2 +12x4 - 12x6) : (2+4+6+....+18+20)

=12 x (2+4-6) : (2+4+6+...+18+20)

=12 x 0 : (2+4+6+...+18+20)

= 0

0 nha bn

 = ( 12 . 2 + 12 . 4 - 12 . 6 ) : ( 2+4+6+8+...+18+20)

= 12 . ( 2+4-6 ) : ( 2+4+6+8+...+18+20 )

= 12 . 0 : ( 2+4+6+8+...+18+20 )

= 0

thích cho mình nha !

Ta có : \(4^x+12.2^x+32=0\)

<=> \(\left(2^x\right)^2+2.6.2^x+36-4=0\)

<=> \(\left(2^x+6\right)^2-4=0\)

<=> \(\left(2^x+6+2\right)\left(2^x+6-2\right)=0\)

<=> \(\left(2^x+8\right)\left(2^x+4\right)=0\)

<=> \(\left[{}\begin{matrix}2^x+8=0\\2^x+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}2^x=-8\\2^x=-8\end{matrix}\right.\) ( Vô lý )

Vậy phương trình vô nghiệm .