Đáp án:

`a)`

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

` x^2 - x - x^2 +2x = 5`

` (x^2 -x^2) + (2x-x) = 5`

` x = 5`

`b)`

` 4x^3 -36x = 0`

` => 4x(x^2 -9) = 0`

` => 4x(x-3)(x+3) =0`

 `=> 4x = 0` hoặc ` x -3 = 0` hoặc ` x +3 = 0`

` => x ∈ {0 ; 3 ; -3}`

`c)`

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

` 2x^2 -2x = x^2 -2x +1`

` 2x^2 -2x -x^2 +2x -1 = 0`

` x^2 - 1 = 0`

` (x-1)(x+1) = 0`

` => x -1 = 0` hoặc ` x+1 =  0`

` => x= 1` hoặc ` x =-1`

`d)`

` => (x-7)(x-2)(x^2 -9x +20) = 72`

` => (x^2 -9x + 14)(x^2 -9x +20) = 72`

Đặt ` x^2 - 9x +17 = t` ta có

` (t-3)(t+3) = 72`

` => t^2 - 9 = 72`

` => t^2 = 81`

` => t = 9` hoặc ` t = -9`

Với ` t = 9`

` => x^2 -9x + 17 = 9`

` => x^2 -9x +8 = 0`

` => (x-1)(x-8) = 0`

` =>`

\(\left[ \begin{array}{l}x=1\\x=8\end{array} \right.\) 

Với ` t= -9`

` => x^2 - 9x + 17 = -9`

` => x^2 -9x + 26 = 0`

` => (x^2 -9x + 81/4) + 23/4 = 0`

` => (x - 9/2)^2 +23/4 = 0`

Ta có ` (x-9/2)^2 \ge 0`

 `=> (x-9/2)^2  + 23/4 \ge 23/4 > 0`

` =>` Không có `x`

Vậy ` x ∈{1;8}`

`a) x(x-1)-x^2+2x=5`

`<=> x^2-x-x^2+2x=5`

`<=> x=5`

`b) 4x^3-36x=0`

`<=> 4x(x^2-9)=0`

`<=> 4x(x-3)(x+3)=0`

`<=>`\(\left[ \begin{array}{l}x=0\\x-3=0\\x+3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=0\\x=3\\x=-3\end{array} \right.\) 

`c) 2x^2-2x=(x-1)^2`

`<=> 2x^2-2x=x^2-2x+1`

`<=> 2x^2-2x-x^2+2x-1=0`

`<=> x^2-1=0`

`<=> (x-1)(x+1)=0`

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

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

`d) (x-7)(x^2-9x+20)(x-2)=72`

`<=> [(x-7)(x-2)](x^2-9x+20)=72`

`<=> (x^2-9x+14)(x^2-9x+20)-72=0`

`<=> (x^2-9x+17-3)(x^2-9x+17+3)-72=0`

`<=> (x^2-9x+17)^2-9-72=0`

`<=> (x^2-9x+17)^2-81=0`

`<=> (x^2-9x+17-9)(x^2-9x+17+9)=0`

`<=> (x^2-9x+8)(x^2-9x+26)=0`

`<=> (x^2-8x-x+8)(x^2-9x+26)=0`

`<=> (x-8)(x-1)(x^2-9x+26)=0`

Do `x^2-9x+26=x^2-2.x. 9/2+81/4+23/4=(x-9/2)^2+23/4>0` với `∀x`

`=>`\(\left[ \begin{array}{l}x-8=0\\x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=8\\x=1\end{array} \right.\)