2^x+ 2^x+1+ 2^x+2 + 2^x+3 = 2^x+ 2^x.2 + 2^x.2² + 2^x.2³ = 2^x(1+ 2 + 2² + 2³) = 2^x.15 = 120 => 2^x =120:15 = 8 => x = 3

2^x + 2^x+1 +2^x+2 + 2^x+3 = 120

Suy ra 2^x .1 + 2^x . 2 + 2^x . 2² + 2^x . 2³ = 120

Suy ra 2^x .( 1 + 2 + 4 + 8 ) = 120

Suy ra 2^x . 15 = 120

Suy ra 2^x = 120 : 15

Suy ra 2^x = 8

Suy ra 2^x = 2³

Do đó x = 3

Vậy x = 3

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

\(\Rightarrow2^x+2^x.2+2^x.2^2+2^x.2^3=120\)

\(\Rightarrow2^x+2^x.2+2^x.4+2^x.8=120\)

\(\Rightarrow2^x\left(1+2+4+8\right)=120\)

\(\Rightarrow2^x.15=120\)

\(\Rightarrow2^x=120:15\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

a) 1 + 2 + 3 + .... + x = 120 

<=> x(x + 1) : 2 = 120

<=> x(x + 1) = 240

<=> x(x + 1) = 15.16

<=> x = 15 (Vì x \(\inℕ\)) 

b) (x + 1) + (x + 2) + (x + 3)  + .... + (x + 100) = 12650 

=> (x + x + x + .... + x) + (1 + 2 + 3 + .... +100) = 12650 (100 hạng tử x)

=> 100x + 100.(100 + 1) : 2 = 12650

=> 100x + 5050 = 12650

<=> 100x = 7600

<=> x = 76

Vậy x = 76 là giá trị cần tìm 

2^x+2^x+1+2^x+2+2^x+3=120

=>2^x(1+2+2²+2³)=120

=>2^x.15=120

=>2^x=120:15

=>2^x=8

=>2^x=2³

=>x=3

Vậy x=3

2^x + 2^x+1 + 2^x+2 + 2^x+3 = 120

2^x(1+2+2²+2³)=120

2^x.15=120

2^x=8

2^x=2³

=>x=3

<=>8.2^x+4.2^x+2.2^x+2^x=120

<=>15.2^x=120

<=>2^x=8

<=>x=3

2^x+3+2^x+2+2^x+1+2^x=120

2^x(2^3+2^2+2+1)=120

2^x.15=120

2^x=8

2^x=2^3

=> x=3

Bài 2

a) x=5

b)x=1

(10 - 4.x) + 120 : 8 = 5 mũ 2

 (10-4.x) +120 : 8 = 25

 (10-4.x)+120 = 25.8

(10-4.x)=200

4.x=200+10

x=210:4

x=52,5

sai rồi bạn ơi

(10-4.x) + 120 : 2 mũ 3 = 1+4 mũ 2

<=> (10-4.x) + 120 : 8 = 5 mũ 2

<=> ( 10-4.x) + 120 :8 = 25 

<=> (10-4.x ) + 15 = 25

<=> (10-4.x) + 15 = 25

<=> 10-4.x = 25-15

<=> 10-4.x =10

<=> 4.x = 10-10

<=> 4.x=0

x = 0 :4

x= 0

Vậy x=0

Mình học lớp 6 nhé , làm quen đc ko , tớ ở BG