Question

( x - 2 )^2= 1

(2x - 1)^3 = -8

(x - 1)^x+2 = (x - 1)^x+4

(2x-3)^5 = -243

Answer 1

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

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

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

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

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

Vậy \(x\in\left\{3;1\right\}.\)

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

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

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

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

=> \(2x=-1\)

=> \(x=\left(-1\right):2\)

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

Vậy \(x=-\frac{1}{2}.\)

\(\left(2x-3\right)^5=-243\)

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

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

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

=> \(2x=0\)

=> \(x=0:2\)

=> \(x=0\)

Vậy \(x=0.\)

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Answer 2

a) (x - 2)² = 1

=> x - 2 = 1 hoặc x - 2 = -1

x = 3 ; x = 1

Vậy x = 3; x = 1

b) (2x - 1)³ = -8

=> 2x - 1 = -2

2x = -1

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

Vậy x = \(\frac{-1}{2}\)

c) (2x - 3)⁵ = -243

=> (2x - 3)⁵ = (-3)⁵

=> 2x - 3 = -3

2x = 0

x = 0

Vậy x = 0