Lời giải chi tiết:

`a)` `(3x-2)^3=2.32`

`=>(3x-2)^3=64`

`=>(3x-2)^3=4^3`

`=>3x-2=4`

`=>3x=4+2`

`=>3x=6`

`=>x=6:3`

`=>x=2`

Vậy `x=2`

`b)` `(x-5)^7=(x-5)^9` `(x\ge5)`

`=>(x-5)^9-(x-5)^7=0`

`=>(x-5)^7.[(x-5)^2-1]=0`

`=>`\(\left[ \begin{array}{l}(x-5)^7=0\\(x-5)^2-1=0\end{array} \right.\)

`TH1:`

`(x-5)^7=0`

`=>x-5=0`

`=>x=0+5`

`=>x=5`

`TH2:`

`(x-5)^2-1=0`

`=>(x-5)^2=1`

`=>`\(\left[ \begin{array}{l}x-5=1\\x-5=-1\end{array} \right.\)

`=>`\(\left[ \begin{array}{l}x=6\\x=4\end{array} \right.\)

`=>x∈{5;6;4}`

Mà `x\ge5` nên `x∈{5;6}`

Vậy `x∈{5;6}`

Đáp án+Giải thích các bước giải:

 `a) (3x-2)^3=2.32`

`<=> (3x-2)^3=64`

`<=> (3x-2)^3=4^3`

`<=> 3x-2=4`

`<=> 3x=4+2`

`<=> 3x=6`

`<=> x=6:3`

`<=> x=2`

Vậy `x=2`

`b) (x-5)^7=(x-5)^9`

`<=> (x-5)^7-(x-5)^9=0`

`<=> (x-5)^7 -(x-5)^7 . (x-5)^2=0`

`<=> (x-5)^7 [1-(x-5)^2]=0`

`<=> [((x-5)^7=0),(1-(x-5)^2=0):}`

`<=> [(x-5=0),((x-5)^2=1):}`

`<=> [(x=5),((x-5)^2=1^2=(-1)^2):}`

`<=> [(x=5),(x \in {4;6}):}`

Mà `x>=5`

Nên `x \in {6;5}`

Vậy `x \in {5;6}`