Bài `2`:
a) `2x^3-2x^2y=2x^2(x-y)`
b) `(a-1)b-2a(a-1)=(a-1)(b-2a)`
c) `10x^2+10xy-5x-5y`
`=(10x^2+10xy)-(5x+5y)`
`=10x(x+y)-5(x+y)`
`=(x+y)(10x-5)`
`=5(x+y)(2x-1)`
d) `x^2+2xy+y^2-4`
`=(x^2+2xy+y^2)-2^2`
`=(x+y)^2-2^2`
`=(x+y-2)(x+y+2)`
Bài `3`:
a) `(x-1)(x^2+x+1)-x(x^2+1)=4`
⇔`x^3-1-x^3-x=4`
⇔`-x=4+1`
⇔`-x=5`
⇔`x=-5`
Vậy `S={-5}`
b) `15x^2-5x=0`
⇔`5x(3x-1)=0`
⇔\(\left[ \begin{array}{l}5x=0\\3x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=\frac{1}{3}\end{array} \right.\)
Vậy `S={0,1/3}`
c) `(x-2)^2-3(x-2)=0`
⇔`(x-2)(x-2-3)=0`
⇔`(x-2)(x-5)=0`
⇔\(\left[ \begin{array}{l}x-2=0\\x-5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
Vậy `S={2,5}`
d) `x^2-9x^2-x+9=0`
⇔`(x^3-9x^2)-(x-9)=0`
⇔`x^2(x-9)-(x-9)=0`
⇔`(x-9)(x^2-1)=0`
⇔`(x-9)(x-1)(x+1)=0`
⇔\(\left[ \begin{array}{l}x-9=0\\x-1=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=9\\x=1\\x=-1\end{array} \right.\)
Vậy `S={9,1,-1}`
