`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=65\)

<=>  \(5x+15=65\)

<=> \(5x=50\)

<=> \(x=10\)

Vậy...

p/s: chúc bạn học tốt

5x+(1+2+3+4+5)=65

5x+15=65

5x=50

x=10

5x+1+2+3+4+5=65

=>5x+15=65

=>5x=50

=>x=10

\(5x+\frac{5.6}{2}=65\Rightarrow x=\frac{65-15}{5}=10\)

(3/8 + x * 3/2 ) : 65/4 = 1

3/8 + x* 3/2 = 1 * 65/4

3/8 + x* 3/2 = 65/4

x* 3/2 = 65/4 - 3/8

x* 3/2 = 127/8

x =  127/8 : 3/2

x = 127/12

1, 4\(^{x+1}\) + 4\(^0\) = 65

\(\Rightarrow\)4\(^{x+1}\) = 65 - 1

\(\Rightarrow\)x + 1      = 64 : 4

\(\Rightarrow\)x + 1      =  16

\(\Rightarrow\)x = 15

2) 10 + 2x = 16\(^{^2}\): 4\(^3\)

\(\Rightarrow\)10 + 2x = 4

\(\Rightarrow\)2x = 4 - 10

\(\Rightarrow\)2x = -6

\(\Rightarrow\)x = -3

Bài 1

a) 4^x+1 + 4⁰ = 65

=> 4^x+1 + 1 = 65

=> 4^x+1 = 64

=> 4^x+1 = 4³

=>x + 1 = 3

=> x = 2

b) 10 + 2x = 16²: 4³

=> 10 + 2x = ( 4² )² : 4³

=> 10 + 2x = 4⁴ : 4³

=> 10 + 2x = 4

=> 2x = -6

=> x = -3

c) 65 - 4^x+2 = 2018⁰

=> 65 - 4^x+2 = 1

=> 4^x+2 = 64

=> 4^x+2 = 4³

=> x + 2 = 3

=> x = 1

d) 2 < |x| < 4

=> x thuộc { -3 ; -4 }