a) \(\left(x-\frac{1}{2}\right)^3=27\)

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

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

=> \(x=3+\frac{1}{2}\)

=> \(x=\frac{7}{2}\)

Vậy \(x=\frac{7}{2}.\)

b) \(\left(2x-1\right)^3=-27\)

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

=> \(2x-1=-3\)

=> \(2x=\left(-3\right)+1\)

=> \(2x=-2\)

=> \(x=\left(-2\right):2\)

=> \(x=-1\)

Vậy \(x=-1.\)

Chúc bạn học tốt!

( x +2 ) . 3 = 60 

=> x + 2   = 60 : 3 

=> x + 2   = 20 

=> x         = 20 - 2 

=> x         = 18 

... 

(x+2).3=60                         (x-4)^3 = 27

x+2     =60:3                      (x-4)^3 = 3^3         

x+2     = 20                        x-4      = 3

x         =20-2                       x        = 3 +4

x         = 18                         x        = 7

10+2x= 2^4

 10+2x=16

      2x=16-10

      2x= 6

      x  =6 : 2

      x = 3

( x +2 ) . 3 = 60 

=> x + 2   = 60 : 3 

=> x + 2   = 20 

=> x         = 20 - 2 

=> x         = 18 

... 

Bằng nhau

\(\frac{-3.7^4+7^3}{7^5.6-7^3.2}=\frac{7^3\left(-3.7+1\right)}{7^3\left(7^2.6-2\right)}=\frac{-21+1}{49.6-2=4}=\frac{-20}{290}\)

a) \(|2x+1|-19=-7\)

\(\Leftrightarrow|2x+1|=12\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=12\\2x+1=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-13\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-13}{2}\end{cases}}}\)

b) \(-28-7\cdot|-3x+15|=-70\)

\(\Leftrightarrow-7\cdot|-3x+15|=-42\)

\(\Leftrightarrow|-3x+15|=6\)

\(\Leftrightarrow\orbr{\begin{cases}-3x+15=6\\-3x+15=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}-3x=-9\\-3x=-21\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)

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

=> x+2=2x

=> 2=2x-x

=>x=2

tương tự

\(1.\left(x^3-1\right)\left(x^2+1\right)=0\)

\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)

<=>x=1

vậy ...

\(2.\left(2x+6\right)\left(3x^2-12\right)=0\)

\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

vậy ...

Trong Th này bn nên dùng dấu ''hoặc''

a,\(\left(x^3-1\right)\left(x^2+1\right)=0\)

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

\(\left[{}\begin{matrix}x^3=1\\x^2=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

b, \(\left(2x+6\right)\left(3x^2-12\right)=0\)

\(\left[{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x=\pm2\end{matrix}\right.\)