-1 . \(\dfrac{5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = -\(\dfrac{24}{27}\)

\(\Rightarrow\) \(\dfrac{-5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-24}{27}\)

\(\Rightarrow\) \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-5}{27}\) - \(\dfrac{-24}{27}\) = \(\dfrac{19}{27}\)

\(\Rightarrow\) \(3^3\) . \(\dfrac{7^3}{3^6}\) = \(\dfrac{19}{3^3}\)

\(\Rightarrow\) \(\dfrac{x^3.7^3}{3^3}\) = \(\dfrac{19}{3^3}\)

\(\Rightarrow\) \(x^3\) . \(7^3\) = 19

a) \(\left(2x+1\right)^3=27\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow x=1\)

b) \(\left(2x-1\right)^3=125\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Leftrightarrow x+1=2x\)

\(\Leftrightarrow x=1\)

d) \(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)

a. ( 2x + 1 )³ = 27

<=> ( 2x + 1 )³ = 3³

<=> 2x + 1 = 3

<=> 2x = 2

<=> x = 1

b. ( 2x - 1 )³ = 125

<=> ( 2x - 1 )³ = 5³

<=> 2x - 1 = 5

<=> 2x = 6

<=> x = 3

c. ( x + 1 )⁴ = 2x⁴

<=> x + 1 = 2x

<=> x = 1

d. ( 2x - 1 )⁵ = x⁵

<=> 2x - 1 = x

<=> x = 1

a) \(\left(2x+1\right)^3=27\)

\(\left(2x+1\right)^3=3^3\)

\(\Rightarrow2x+1=3\)

\(2x=3-1\)

\(2x=2\)

\(x=2\div2\)

\(x=1\)

Vậy \(x=1\).

b) \(\left(2x-1\right)^3=125\)

\(\left(2x-1\right)^3=5^3\)

\(\Rightarrow2x-1=5\)

\(2x=5+1\)

\(2x=6\)

\(x=6\div2\)

\(x=3\)

Vậy \(x=3\).

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Rightarrow x+1=2x\)

\(x-2x=-1\)

\(-x=-1\)

\(x=1\)

Vậy \(x=1\).

d) \(\left(2x-1\right)^5=x^5\)

\(\Rightarrow2x-1=x\)

\(2x-x=1\)

\(x=1\)

Vậy \(x=1\).

\(\frac{1}{9}\cdot27^x=3^x\)

\(3^x:27^x=\frac{1}{9}\)

\(\left(3:27\right)^x=\frac{1}{9}\)

\(\left(\frac{1}{9}\right)^x=\frac{1}{9}\)

\(\Rightarrow x=1\)

Câu này lớp mình mới học hồi thứ hai.

\(3^{x-2}=27\)

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

=>\(x-2=3\)

=>\(x=5\)

<=>x-2=3

<=>x=5

\(3^{x-2}=27\\ \Leftrightarrow3^{x-2}=3^3\)

Ta có : \(x-2=3\\ \Leftrightarrow x=2+3\\\Leftrightarrow x=5 \)

có ai giải giúp mình ko mình đang gấp huhu

2^x x 16 = 128 

2^x          = 128 : 16

2 ^x        = 8

2^x         = 2³

3^x : 9 = 27 

3^x      = 27 x 9

3^x      =243

3^x      = 3⁵

[ 2^x + 1 ]³ = 27

2^x3 + 1³   = 27

2^x3 +1      = 27

2^x3           = 27 - 1

2^x3           = 26

a, (-0,2)² \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)

= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)

= 0,2 - \(\dfrac{2}{3}\)

= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)

=  - \(\dfrac{7}{15}\)

b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)

 = \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)

= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)

= \(\dfrac{1}{2}\)

a, (-0,2)2 $\times$ 5 - 213×27346×9546×95213×273​

= 0,04 $\times$ 5 - 213×39212×310212×310213×39​

= 0,2 - 2332​

= 210102​ - 2332​

=  - 715157​

b, 56+22.253+23.125226.5626.5656+22.253+23.1252​

 = 56+4.56+8.5626.5626.5656+4.56+8.56​

= 56.(1+4+8)26.5626.5656.(1+4+8)​

= 1221​
 

125:(2x-3)=25

-->2x-3=125:25

-->2x-3=5

-->2x=5+3

-->2x=8

-->x=8:2

-->x=4

k nha

\(125:\left(2x-3\right)=5^5:5^3\)

\(125:\left(2x-3\right)=5^2=25\)

\(2x-3=5\)

\(2x=8\)

\(x=4\)

125:(2x-3)=25

2x-3=125:25

2x-3=5

2x=5+3

2x=8

x=8:2

x=4

x¹³ = 27.x¹⁶

=> x¹³ - 27x¹⁶ = 0

=> x¹³(1 - 27x³) = 0

=> \(\orbr{\begin{cases}x^{13}=0\\1-27x^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\27x^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^3=\frac{1}{27}\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

c) \(\left(\frac{1}{2}\right)^{2x-1}=\frac{1}{8}\)

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

=> \(2x-1=3\)

=> \(2x=3+1\)

=> \(2x=4\)

=> \(x=4:2=2\)

\(b,\text{ }x^{13}=27\cdot x^{16}\)

\(x^{16}\text{ : }x^{13}=27\)

\(x^3=3^3\)

\(x=3\)

\(b,\text{ }x^{13}=27\cdot x^{16}\)

\(x^{16}\text{ : }x^{13}=27\)

\(x^3=3^3\)

\(x=3\)

a. ( 2x + 1 )² = 49

<=> ( 2x + 1 )² = 7²

<=> 2x + 1 = 7

<=> x = 3

b. ( 2x - 1 )⁴ = 81

<=> ( 2x - 1 )⁴ = 3⁴

<=> 2x - 1 = 3

<=> x = 2

c. ( x + 1 )³ = 2x³

<=> x + 1 = 2x

<=> x = 1

d. ( 2x + 1 )³ = 3x³

<=> 2x + 1 = 3x

<=> x = 1

( 2x + 1 )² = 49

<=> ( 2x + 1 )² = ( ±7 )²

<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

( 2x - 1 )⁴ = 81

<=> ( 2x - 1 )⁴ = ( ±3 )⁴

<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

( x + 1 )³ = ( 2x )³

<=> x + 1 = 2x

<=> x - 2x = -1

<=> -x = -1

<=> x = 1

( 2x + 1 )³ = ( 3x )³

<=> 2x + 1 = 3x

<=> 2x - 3x = -1

<=> -x = -1

<=>  x = 1

a) \(\left(2x+1\right)^2=49\)  

\(\left(2x+1\right)=\pm7\) 

\(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\)  

\(\orbr{\begin{cases}2x=6\\2x=-8\end{cases}}\)

\(\orbr{\begin{cases}x=3\\x=-4\end{cases}}\) 

b) \(\left(2x-1\right)^4=81\) 

\(\left(2x-1\right)^4=3^4\) 

\(2x-1=\pm3\) 

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\) 

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\) 

\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)    

c) \(\left(x+1\right)^3=\left(2x\right)^3\)    

\(x+1=2x\) 

\(x-2x=-1\) 

\(-x=-1\) 

\(x=1\)      

d) \(\left(2x+1\right)^3=\left(3x\right)^3\)     

\(2x+1=3x\) 

\(1=3x-2x\) 

\(1=x\)