Đáp án:

Bài `1`:

`a. x^2-6x+8=0`

`<=> x^2-4x-2x+8=0`

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

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

`<=>`\(\left[ \begin{array}{l}x-4=0\\x-2=0\end{array} \right.\) 

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

Vậy `S={2;4}`

`b. 9x^2+6x-8=0`

`<=> 9x^2+12x-6x-8=0`

`<=> (3x+4)(3x-2)=0`

`<=>`\(\left[ \begin{array}{l}3x+4=0\\3x-2=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-\dfrac{4}{3}\\x=\dfrac{2}{3}\end{array} \right.\) 

Vậy `S={-4/3;2/3}`

`c. x^3+x^2+x+1=0`

`<=> x^2(x+1)+(x+1)=0`

`<=> (x+1)(x^2+1)=0`

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

`<=>`\(\left[ \begin{array}{l}x=-1\\x^2=-1\ (\text{vô nghiệm, }x^2\ge 0\forall x)\end{array} \right.\) 

Vậy `S={-1}`

`d. x^3-x^2-x+1=0`

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

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

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

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

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

Vậy `S={+-1}`

Bài `2`:

`a. x^3+3x^2+x+3xy^2+y+y^3`

`= (x+y)^3+(x+y)`

`= (x+y)[(x+y)^2+1]`

`= (x+y)(x^2+2xy+y^2+1)`

`b.x^3+y(1-3x^2)+x(3y^2-1)-y^3`

`= x^3+y-3x^2y+3xy^2-x-y^3`

`= (x^3-3x^2y+3xy^2-y^3)-(x-y)`

`= (x-y)^3-(x-y)`

`= (x-y)[(x-y)^2-1]`

`= (x-y)(x^2-2xy+y^2-1]`

`c. 27x^3+27x^2+9x+1`

`= (3x+1)^3`

`d. x^2y+xy^2-x-y`

`= xy(x+y)-(x+y)`

`= (x+y)(xy-1)`

`e. 8xy^3-5xyz-24y^2+15z`

`= 8xy^3-24y^2-5xyz+15z`

`= 8y^2(xy-3)-5z(xy-3)`

`= (xy-3)(8y^2-5z)`

Bài `3`:

`a. x^2+6x+9`

`= (x+3)^2`

`b. 10x-25-x^2`

`= -(x^2-10x+25)`

`= -(x-5)^2`

`c. (a+b)^2+(a-b)^3`

`= (a+b+a-b)[(a+b)^2-(a+b)(a-b)+(a-b)^2]`

`= 2a[a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2]`

`= 2a[a^2+3b^2]`

`d. (a+b)^2-(a-b)^3`

`= (a+b-a+b)[(a+b)^2+(a+b)(a-b)+(a-b)^2]`

`= 2b[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2]`

`= 2b[3a^2+b^2]`

`e. x^3+27`

`= x^3+3^3`

`= (x+3)(x^2-3x+9)`