a, 2^x.16 = 1024 => 2^x = 1024:16 => 2^x = 64 => 2^x = 2⁶ => x = 6

b, x¹⁷ = x

=> x¹⁷ - x = 0

=> x(x¹⁶-1)=0

=> x = 0 hoặc x¹⁶ - 1 = 0

=> x = 0 hoặc x¹⁶ = 1

=> x = 0 hoặc x = 1

c, (2x-2)³=64

=> (2x-2)³ = 4³

=>2x-2=4

=>2x=6

=>X=3

d,(x-6)² = (x-6)³

=> (x-6)²-(x-6)³=0

=> (x-6)²-[1-(x-6)] = 0

=> (x-6)² = 0 hoặc 1 - (x-6) = 0

=> x - 6 = 0 hoặc x - 6 = 1

=> x = 6 hoặc x = 7

e, 3 + 2^x-1 = 24-[4²-(2²-1)]

=> 3 + 2^x-1 = 11

=> 2^x-1 = 8

=> 2^x-1 = 2³

=>x-1=3

=>x=4

a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15

=>x=-1/3:1/15=5

b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)

=>x*2/3=-1

=>x=-3/2

c: =>2x+1=4 hoặc 2x+1=-4

=>x=3/2 hoặc x=-5/2

h: =>x-3=4

=>x=7

g: =>2x-1=3

=>2x=4

=>x=2

f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)

=>x*-5/6=5/6

=>x=-1

d: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>x=-1 hoặc x=2

a: \(2^{x^2-1}=256\)

=>\(2^{x^2-1}=2^8\)

=>\(x^2-1=8\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

b: \(3^{x^2+3x}=81\)

=>\(3^{x^2+3x}=3^4\)

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

c: \(2^{x^2-5x}=64\)

=>\(2^{x^2-5x}=2^6\)

=>\(x^2-5x=6\)

=>\(x^2-5x-6=0\)

=>(x-6)(x+1)=0

=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^x=243\)

=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)

=>x=-5

e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)

=>\(3^{-x-5}=3^{2x+1}\)

=>-x-5=2x+1

=>-3x=6

=>x=-2

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) \(3^2.x+2^3.x=51\)

\(\Leftrightarrow x\left(3^2+2^3\right)=51\)

\(\Leftrightarrow17x=51\)

\(\Leftrightarrow x=3\)

Vậy

b) \(6^2.2-\left(84-3^2.x\right):7=69\)

\(\Leftrightarrow\left(84-3^2.x\right):7=3\)

\(\Leftrightarrow84-3^2.x=21\)

\(\Leftrightarrow3^2.x=63\)

\(\Leftrightarrow x=7\)

Vậy

c) \(\left(x+1\right)^2=64\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=8\\x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-9\end{cases}}}\)

Vậy

a ) 9 . 3x = 81

          3x = 9

            x = 3

b ) 2x : 4 = 1

     2x      = 4

       x      = 2

c ) 2x - 64 = 2

     2x        = 66

       x        = 33

d ) 2x = 16

       x  = 8

e ) 3 ^ 2 . 3 ^ 4 . 3x = 3 ^ 10

     3 ^ ( 2 + 4 + x )  = 3 ^ 10

=> 2 + 4 + x = 10

                 x = 4

f ) 2x + 4 . 2x = 5 . 2 ^ 5

    2x + 8 x     = 160

    10x            = 160

        x            = 16

\(a.9.3x=81\)

\(3x=81:9\)

\(3x=9\)

\(x=9:3\)

\(x=3\)

a,

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\\ \)

\(\dfrac{1}{4}:x=\dfrac{8-15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

x = \(\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

b,

( 3x + 1)^3 = 64

(3x + 1)^3 = 4^3

(3x + 1) = 4

3x = 4 - 1

3x = 3

x = 3 : 3

x = 1

c,

( 2x - 3)^4 = 81

( 2x - 3) ^4 = 3^4

(2x - 3) = 3

2x = 3 + 3

2x = 6

x = 6: 2

x = 3