Để phép nhân có giá trị=0 thì 1 trong 2 thừa số phải=0

⇒x=0 hoặc x+17=0

   x=0 hoặc x       =-17

Vậy x ϵ (0;-17)

4.2\(^x\)+3.2\(^{x+1}\)=80

4.2\(^x\)+3.2\(^x\).2\(^1\)=80

2\(^x\).(4+3.2)=80

2\(^x\).10=80

2\(^x\)=8

⇒x=3 vì 2\(^3\)=8

= 2² .2^X+3.2X.2¹=80

=4.2^X+3.2.2^X

= 4.2^X+6.2x

=2X.(4+6)

= 2X.10=80

2X=80:10

2X= 8

X= 8:2

X=4

( x + 2 ) x (x - 3 ) = 5

X x (2 +3 ) = 5

X x 5 =5

     x= 5:5

     x= 1

\(5\cdot\left(x-2\right)^3-3\cdot\left(x-2\right)^3=250\)

\(\left(x-2\right)^3\cdot\left(5-3\right)=250\)

\(\left(x-2\right)^3\cdot2=250\)

\(\left(x-2\right)^3=\dfrac{250}{2}\)

\(\left(x-2\right)^3=125\)

\(\left(x-2\right)^3=5^3\)

\(x-2=5\)

\(x=5+2\)

\(x=7\)

\(5.\left(x-2\right)^3-3\left(x-2\right)^3=250\)

\(\Leftrightarrow\left(x-2\right)^3.\left(5-3\right)=250\)

\(\Leftrightarrow\left(x-2\right)^3.2=250\)

\(\Leftrightarrow\left(x-2\right)^3=125\)

\(\Leftrightarrow\left(x-2\right)^3=5^3\)

\(\Leftrightarrow x-2=5\)

\(\Leftrightarrow x=7\)

Ai giúp v

`S=4+3^{2} +3^{3}+...+3^{223}`

`=>3S=12+3^{3}+3^{4}+....+3^{224}`

Có: \(3S-S=12+3^{3}+3^{4}+....+3^{224}-4-3^{2}-3^{3}-...-3^{233}\)

 `=>2S=12+3^{224}-4-3^2`

\(=>S=4+\dfrac{3^{224}-3^{2}}{2}\)