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

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

     \(x-1=\dfrac{1}{2}\)

     \(x=\dfrac{1}{2}+1\)

     \(x=\dfrac{3}{2}\)

(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

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

=> \(x=\pm\dfrac{1}{2}\)

=> x = 0 

vậy x = 0 :))

16x⁴-72x²+90=9

=> 16x⁴-72x²+81=0

=> (9-4x²)²=0

\(\dfrac{1}{6}+x=\dfrac{5}{12}\)
\(=>x=\dfrac{5}{12}-\dfrac{2}{12}=\dfrac{1}{4}\)
\(\dfrac{3}{4}+\dfrac{1}{4}x=-\dfrac{1}{2}\)
\(=>\dfrac{1}{4}x=-\dfrac{5}{4}\)
\(=>x=-\dfrac{5}{4}.4=-5\)
\(7^{2x}+7^{2x+3}=344\)
\(< =>49^x+49^x.343=344\)
\(=>x=?\)