`1)`

 $ (x ^ 2 - 5x ) ^ 2 + 10 (x ^ 2 - 5x) + 24 = 0 \\\Leftrightarrow (x ^ 2 - 5x + 6) (x ^ 2 - 5x + 4) = 0 \\\Leftrightarrow \left[\begin{array}{l}x ^ 2 - 5x + 6 = 0 \\  x ^ 2 - 5x + 4 = 0\end{array}\right. \\\Leftrightarrow \left[\begin{array}{l}x = 3 \\ x = 2 \\ x = 1 \\ x = 4\end{array}\right. $

 `2)`

Đặt `A = x2+x+1` thay vào phương trình ta được:

`A.(A+1) = 12`

`<=> A2 + A - 12 = 0`

`<=> A2 + 4A - 3A - 12 = 0`

`<=> A.(A+4) - 3.(A+4) =0`

`<=> (A-3).(A+4) = 0` `(1)`

Thay `A =  x2+x+1` vào `(1)` ta được:

`( x2+x-2).( x2+x+5) = 0`

`<=> (x2+2x-x-2).(x2+x+5) = 0`

`<=> [x(x+2)-(x+2)].(x2+x+5) = 0`

`<=> (x-1)(x+2)(x2+x+5) = 0`

`<=>`\(\left[ \begin{array}{l}x-1=0\\x+2=0\\x2+x+5=0\end{array} \right.\) 

TH`1:`

`x-1 =0`

`<=> x = 1`

TH `2:` 

`x+2 = 0`

`<=> x = -2`

TH`3:`

`x2+x+5 =0` `(2)`

`<=> x2 + 2.x.1/2 + (1/2)2 + 19/4 = 0 ` 

`<=> (x+1/2)2 +19/4 = 0`

Vì `(x+1/2)2 \ge 0` với mọi `x`

`=> (x+1/2)2 +19/4 \ne 0` với mọi `x`

Nên `(2)` vô lí

`S={-2;1}`  

`3)`

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt \(x^2+3x+1=t\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-24=0\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)

TH1:t=5\(\Rightarrow x^2+3x+1=5\)

\(\Leftrightarrow x^2+3x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

TH2:t=-5\(\Rightarrow x^2+3x+1=-5\)

\(\Leftrightarrow x^2+3x+6=0\)(vô nghiệm)

`4)`

\(\left(x+1\right)^3+\left(x-2\right)^3=\left(2x-1\right)^3\\ \Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=8x^3-12x^2+6x-1\\ \Leftrightarrow2x^3-3x^2+15x-7-8x^3+12x^2-6x+1=0 \)\(\Leftrightarrow-6x^3+9x^2+9x-6=0\\ \Leftrightarrow-3\left(2x^3-3x^2-3x+2\right)=0\\ \Leftrightarrow\left(2x^3+2\right)-\left(3x^2+3x\right)=0\)

\(\Leftrightarrow2\left(x^3+1\right)-3x\left(x+1\right)=0\\ \Leftrightarrow2\left(x^2-x+1\right)\left(x+1\right)-3x\left(x+1\right)=0\\ \Leftrightarrow\left(2x^2-2x+2-3x\right)\left(x+1\right)=0\\ \Leftrightarrow\left(2x^2-5x+2\right)\left(x+1\right)=0\\ \Leftrightarrow\left(2x^2-4x-x+2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[\left(2x^2-4x\right)-\left(x-2\right)\right]\left(x+1\right)=0\\ \Leftrightarrow\left[2x\left(x-2\right)-\left(x-2\right)\right]\left(x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\\x=-1\end{matrix}\right.\)

\(S=\dfrac{1}{2};2;-1\)