`text{Đáp án:}`

`1.`  \(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\) 

`2.`  \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\) 

`3.`  `x=-5`

`text{Giải thích các bước giải:}`

`1.`
`(x-2)^8=(x-2)^6`
`=>(x-2)^8-(x-2)^6=0`
`=>(x-2)^6.(x-2)^2-(x-2)^6=0`
`=>(x-2)^6.(x-2)^2-(x-2)^6 .1=0`
`=>(x-2)^6.[(x-2)^2-1]=0`
`=>`\(\left[ \begin{array}{l}(x-2)^6=0\\(x-2)^2 -1=0\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}(x-2)=0\\(x-2)^2=0+1\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x-2=0\\(x-2)^2=1\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x=0+2\\(x-2)^2=1^2\\(x-2)^2=(-1)^2\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x=2\\x-2=1\\x-2=-1\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x=2\\x=1+2\\x=-1+2\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\) 
`2.`
`(x+2)^2=36`
`=>`\(\left[ \begin{array}{l}(x+2)^2=6^2\\(x-2)^2=(-6)^2\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x+2=6\\x+2=-6\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x=6-2\\x=-6-2\end{array} \right.\) 
`=>`\(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\) 
`3.`
`(x+2)^3=-27`
`=>(x+2)^3=(-3)^3`
`=>x+2=-3`
`=>x=-3-2`
`=>x=-5`

Đáp án:

1. Ta có : 

`(x - 2)^8 = (x - 2)^6`

` <=> (x - 2)^8  - (x - 2)^6 = 0`

` <=> (x - 2)^6.[(x - 2)^2 - 1] = 0` 

<=> \(\left[ \begin{array}{l}(x - 2)^6 = 0\\(x - 2)^2 - 1 = 0\end{array} \right.\) 

<=> \(\left[ \begin{array}{l}x - 2 = 0\\(x - 2)^2 = 1\end{array} \right.\) 

<=> \(\left[ \begin{array}{l}x=2\\x=3 ; x = 1\end{array} \right.\) 

Vậy `x = 1 ; x = 2 ; x = 3`

2. Ta có : 

`(x + 2)^2 = 36`

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

<=> \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\) 

3. Ta có : 

`(x + 2)^3 = -27`

` <=> (x  + 2)^3 = (-3)^3`

` <=> x + 2 = -3`

` <=> x = -5`

Giải thích các bước giải: