a: =>2^x*4-2^x*3=32

=>2^x=32

=>x=5

b: =>(4x-3)^2-(4x-3)=0

=>(4x-3)(4x-3-1)=0

=>(4x-3)(4x-4)=0

=>x=3/4 hoặc x=1

c: =>7^2x+7^2x*7^3=344

=>7^2x=1

=>2x=0

=>x=0

d: =>(7x-3)^2012-(7x-3)^2010=0

=>(7x-3)^2010*[(7x-3)^2-1]=0

=>(7x-3)^2010*(7x-4)(7x-2)=0

=>x=2/7; x=4/7; x=3/7

e: =>(4x^2-3)^3=-8

=>4x^2-3=-2

=>4x^2=1

=>x^2=1/4

=>x=1/2 hoặc x=-1/2

a) 2^x(2² - 3) = 32

2^x.1=2⁵

=> x = 5

b) (4x - 3)² = 4x -3

=> (4x - 3)² - (4x - 3) = 0

(4x-3)[(4x - 3) - 1] = 0

(4x-3)(4x - 4)=0

\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\4x-4=0\end{matrix}\right.\)         \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\end{matrix}\right.\)

c) 7^2x + 7^2x+3 = 344

=> 7^2x(1 + 7³) =344

7^2x . 344 = 344

=> 2x = 0  => x = 0

d) (7x - 3)²⁰¹² = (3 - 7x)²⁰¹⁰

=> (7x - 3)²⁰¹² - (7x - 3)²⁰¹⁰ = 0

(7x - 3)²⁰¹⁰ [(7x - 3)² - 1] = 0

\(\Rightarrow\left[{}\begin{matrix}7x-3=0\\\left(7x-3\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\7x=4\\7x=2\end{matrix}\right.\)                 \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{4}{7}\\x=\dfrac{2}{7}\end{matrix}\right.\)

e) (4x² - 3)³ + 8 = 0

(4x² - 3)³ = (-2)³

=> 4x² - 3 = -2

4x² = 1

x² = 1/4

=> \(x=\pm\dfrac{1}{2}\)

a: Ta có: \(2\left(x-2\right)^3=2-x\)

\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

b: ta có: \(8x^3-72x=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

c: Ta có: \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

Lời giải:
$7^{2x-6}=49=7^2$

$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

a: \(\Leftrightarrow8x\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{0;3;-3\right\}\)

b: \(\Leftrightarrow x^2-4x+4-x^2-2x+3=12\)

=>-6x=5

hay x=-5/6

(5 - \(x\))(9\(x^2\) - 4) =0

\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}

7^2\(x\)  + 7^{2\(x\) + 3} = 344

7^2\(x\)  \(\times\) ( 1 + 7³) = 344

7^2\(x\)  \(\times\) (1 + 343) = 344

7^2\(x\)  \(\times\) 344        = 344

7^2\(x\)                    = 344 : 344

7^2\(x\)                  = 1

7^2\(x\)                 =  7⁰

\(2x\)                  = 0

\(x\)                   = 0

Kết luận: \(x\) = 0

|2 - 2\(x\)| - 3,75 = (-0,5)²

|2 - 2\(x\)| - 3,75 = 0,25

|2- 2\(x\)|           =0,25 + 3,75

|2 - 2\(x\)|          = 4

\(\left[{}\begin{matrix}2-2x=-4\left(x>1\right)\\2-2x=4\left(x< 1\right)\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=6\left(x\ge1\right)\\2x=-2\left(x\le1\right)\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Kết luận: \(x\) \(\in\) { -1; 3}

a) (x - 1)² = 1.

<=> x - 1 = 1 hoặc x - 1 = -1.

<=> x = 2 hoặc x = 0.      

b) 7^{2x - 6} = 49.

<=> 7^{2x - 6} = 7².   

<=> 2x - 6 = 2.

<=> x = 4.  

c) (2x - 16)⁷ = 128.

<=> (2x - 16)⁷ = 2⁷.

<=> 2x - 16 = 2.

<=> x = 9.

16x⁴-72x²+90=9

=> 16x⁴-72x²+81=0

=> (9-4x²)²=0

b) \(\left(x-1\right)^3=\dfrac{1}{8}\)

    \(\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)

     \(x-1=\dfrac{1}{2}\)

     \(x=\dfrac{1}{2}+1\)

     \(x=\dfrac{3}{2}\)