a) (x - 2)^2 = 16

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

=> x - 2 = 4 hoặc x - 2 = -4

=> x = 4 + 2 = 6

=> x = (-4) + 2 = -2

Vậy x = 6 hoặc x = -2

b) (2x - 1)^3 = 8

(2x - 1)^3 = 2^3

=>2x - 1 = 2

2x = 2 + 1 = 3

\(x=\frac{3}{2}\)

c) (2x - 1)^4 = 81

(2x - 1)^4 = 3^4=(-3)^4

=> 2x - 1 = 3 hoặc 2x - 1 = -3

=>2x = 3 + 1  = 4

=>x = 4 : 2 = 2

2x = (-3) + 1 = -2

x = -1

Vậy x = 2 hoặc x = -1

Mik cũng giống bạn 

Các đề bài trên khi chuyển vế đều bị mất đi x nên không có x thỏa mãn

Giúp mk vs

Bài 2: 

a: Ta có: \(x\left(2x-1\right)-2x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

1/

\(B=\frac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{3^{16}-1-3^{16}}{4}=\frac{-1}{4}\)

2/

a, (x-5)²-(x+3)²=1

<=>(x-5+x+3)(x-5-x-3)=1

<=>-16.(x-1)=1

<=>x-1=-1/16

<=>x=15/16

b, (2x-1)²-(2x-3)²=4

<=>(2x-1+2x-3)(2x-1-2x+3)=4

<=>-8(x-1)=4

<=>x-1=-1/2

<=>x=1/2

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)