a: \(x^3=27\)

=>\(x^3=3^3\)

=>x=3

b: \(\left(2x-1\right)^3=8\)

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

=>2x-1=2

=>2x=2+1=3

=>\(x=\frac32=1,5\)

c: \(\left(x-2\right)^2=16\)

=>\(\left[\begin{array}{l}x-2=4\\ x-2=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4+2=6\\ x=-4+2=-2\end{array}\right.\)

d: \(\left(2x-3\right)^2=9\)

=>\(\left[\begin{array}{l}2x-3=3\\ 2x-3=-3\end{array}\right.\Longrightarrow\left[\begin{array}{l}2x=6\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=0\end{array}\right.\)

e: \(2x+5=3^4:3^2\)

=>\(2x+5=3^2=9\)

=>2x=9-5=4

=>\(x=\frac42=2\)

f: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

a: \(x^3=27\)

=>\(x^3=3^3\)

=>x=3

b: \(\left(2x-1\right)^3=8\)

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

=>2x-1=2

=>2x=2+1=3

=>\(x=\frac32=1,5\)

c: \(\left(x-2\right)^2=16\)

=>\(\left[\begin{array}{l}x-2=4\\ x-2=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4+2=6\\ x=-4+2=-2\end{array}\right.\)

d: \(\left(2x-3\right)^2=9\)

=>\(\left[\begin{array}{l}2x-3=3\\ 2x-3=-3\end{array}\right.\Longrightarrow\left[\begin{array}{l}2x=6\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=0\end{array}\right.\)

e: \(2x+5=3^4:3^2\)

=>\(2x+5=3^2=9\)

=>2x=9-5=4

=>\(x=\frac42=2\)

f: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

`Answer:`

`\frac{x-1}{34}+\frac{x-2}{34}=\frac{x-3}{32}+\frac{x-4}{31}`

`<=>\frac{x-1}{34}-1+\frac{x-2}{33}-1=\frac{x-3}{32}-1+\frac{x-4}{31}-1`

`<=>\frac{x-35}{34}+\frac{x-35}{33}=\frac{x-35}{32}+\frac{x-35}{31}`

`<=>(x-35)(\frac{1}{34}+\frac{1}{33}-\frac{1}{32}-\frac{1}{31})=0`

`<=>x-35=0`

`<=>x=35`

a)  | x | − 3 4 = 5 3 ⇒ | x | = 29 12 ⇒ x = 29 12 x = − 29 12

b)  x − 3 2 = 1 ⇒ x − 3 2 = 1 x − 3 2 = − 1 ⇒ x = 5 2 x = 1 2

a: 3x=81

nên x=27

b: \(5\cdot4^x=80\)

\(\Leftrightarrow4^x=16\)

hay x=2

c: \(2^x=4^5:4^3\)

\(\Leftrightarrow2^x=2^4\)

hay x=4

\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)

\(\left|x+\frac{3}{4}\right|=\frac{1}{3}\)

\(\Rightarrow x+\frac{3}{4}=\pm\frac{1}{3}\)

\(\cdot x+\frac{3}{4}=\frac{1}{3}\)

\(x=-\frac{5}{12}\)

\(\cdot x+\frac{3}{4}=-\frac{1}{3}\)

\(x=-\frac{13}{12}\)

a)x=37

b)x=-28

c)x=-48

d)x=15