Bài 1:
`a) (5x-2y)(5x+2y)+4y-1`
`=(5x)^2-(2y)^2+4y-1`
`=(5x)^2-4y^2+4y-1`
`=(5x)^2-(4y^2-4y+1)`
`=(5x)^2-(2y-1)^2`
`=(5x-2y+1)(5x+2y-1)`
`b) x^2(xy+1)+2y-x-3xy`
`=x^3y+x^2+2y-x-3xy`
`=(x^3y-xy)+(x^2-x)-(2xy-2y)`
`=xy(x^2-1)+x(x-1)-2y(x-1)`
`=xy(x-1)(x+1)+x(x-1)-2y(x-1)`
`=(x-1)(x^2y+xy+x-2y)`
Bài 2:
`a) (x+1/2)^2-(x+1/2)(x+6)=8`
`<=> (x+1/2)(x+1/2-x-6)=8`
`<=> -11/2(x+1/2)=8`
`<=> x+1/2=-16/11`
`<=> x=-43/22`
`b) (x^2+2x)^2-2x^2-4x=3`
`<=>x^4+4x^3+4x^2-2x^2-4x-3=0`
`<=> x^4+4x^3+2x^2-4x-3=0`
`<=> x^4-x^3+5x^3-5x^2+7x^2-7x+3x-3=0`
`<=> x^3(x-1)+5x^2(x-1)+7x(x-1)+3(x-1)=0`
`<=> (x-1)(x^3+5x^2+7x+3)=0`
`<=>(x-1)(x^3+x^2+4x^2+4x+3x+3)=0`
`<=> (x-1)[x^2(x+1)+4x(x+1)+3(x+1)]=0`
`<=> (x-1)(x+1)(x^2+4x+3)=0`
`<=> (x-1)(x+1)(x^2+x+3x+3)=0`
`<=> (x-1)(x+1)^2(x+3)=0`
`<=>`\(\left[ \begin{array}{l}x-1=0\\x+1=0\\x+3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x=-1\\x=-3\end{array} \right.\)
