Đáp án + Giải thích các bước giải:
Bài `1` :
`a)` `4y^2-6y=2y(2y-3)`
`b)` `x^2+6x+9-y^2=(x+3)^2-y^2`
`=(x+3-y)(x+3+y)`
`c)` `2x^2+4xy-3x-6y`
`=2x(x+2y)-3(x+2y)`
`=(x+2y)(2x-3)`
`d)` `x^3-x^2-3x+27`
`=x^3+27-x^2-3x`
`=(x+3)(x^2-3x+9)-x(x+3)`
`=(x+3)(x^2-3x+9-x)`
`=(x+3)(x^2-4x+9)`
Bài `2` :
`a)` `x^2-3x=0`
`<=>x(x-3)=0`
`<=>[(x=0),(x-3=0):}`
`<=>[(x=0),(x=3):}`
Vậy `S={0;3}`
`b)` `2x^2(x-2)+8(2-x)=0`
`<=>2x^2(x-2)-8(x-2)=0`
`<=>(x-2)(2x^2-8)=0`
`<=>(x-2)*2(x^2-4)=0`
`<=>(x-2)*2(x-2)(x+2)=0`
`<=>2(x-2)^2(x+2)=0`
`<=>(x-2)^2(x+2)=0`
`<=>[((x-2)^2=0),(x+2=0):}`
`<=>[(x-2=0),(x=-2):}`
`<=>[(x=2),(x=-2):}`
Vậy `S={2;-2}`
`c)` `(x+1)^2-3x^2-3x=0`
`<=>(x+1)^2-3x(x+1)=0`
`<=>(x+1)[(x+1)-3x]=0`
`<=>(x+1)(x+1-3x)=0`
`<=>(x+1)(1-2x)=0`
`<=>`\(\left[ \begin{array}{l}x+1=0\\1-2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `S={-1;1/2}`
`d)` `x^4+2x^2+1-2(x^2+1)x+x^2=9`
`<=>x^4+2x^2+1-2x(x^2+1)+x^2=9`
`<=>x^4+2x^2+1-2x^3-2x+x^2=9`
`<=>x^4+3x^2+1-2x^3-2x=9`
`<=>x^4+3x^2+1-2x^3-2x-9=0`
`<=>x^4+3x^2-2x^3-2x-8=0`
`<=>(x-2)(x+1)(x^2-x+4)=0`
`<=>[(x-2=0),(x+1=0),(x^2-x+4=0):}`
`<=>[(x=2),(x=-1),(x=emptyset):}`
Vậy `S={2;-1}`.
