Bài 2:
`a)x²-8x`
`=x.x-8.x`
`=x(x-8)`
`b)x²-xy-6x+6y`
`=(x²-xy)-(6x-6y)`
`=x(x-y)-6(x-y)`
`=(x-y)(x-6)`
`c)x²-6x+9-y²`
`=(x²-6x+9)-y²`
`=(x-3)²-y²`
`=(x-3+y)(x-3-y)`
`d)x³+y³+2x+2y`
`=(x³+y³)+(2x+2y)`
`=(x+y)(x²-xy+y²)+2(x+y)`
`=(x+y)(x²-xy+y²+2)`
`e)4y²-x²+4x-4`
`=(2y)²-(x²-4x+4)`
`=(2y)²-(x-2)²`
`=[2y+(x-2)][2y-(x-2)]`
`=(2y+x-2)(2y-x+2)`
Bài 3:
`a)(2x-3)²-49=0`
`⇔(2x-3)²-7²=0`
`⇔(2x-3+7)(2x-3-7)=0`
`⇔(2x+4)(2x-10)=0`
`⇔4(x+2)(x-5)=0`
`⇔(x+2)(x-5)=0`
`⇔`$\left[\begin{matrix} x+2=0\\ x-5=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-2\\ x=5\end{matrix}\right.$
Vậy `x∈{-2;5}`
`b)2x(x-5)-35+7x=0`
`⇔2x(x-5)+(7x-35)=0`
`⇔2x(x-5)+7(x-5)=0`
`⇔(x-5)(2x+7)=0`
`⇔`$\left[\begin{matrix} x-5=0\\ 2x+7=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=5\\ x=-\dfrac{7}{2}\end{matrix}\right.$
Vậy `x∈{5;-7/2}`
`c)x²-3x-10=0`
`⇔x²-5x+2x-10=0`
`⇔x(x-5)+2(x-5)=0`
`⇔(x+2)(x-5)=0`
`⇔`$\left[\begin{matrix} x+2=0\\ x-5=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-2\\ x=5\end{matrix}\right.$
Vậy `x∈{-2;5}`
`d)4x(2-x)+(2x+1)²=2`
`⇔8x-4x²+4x²+4x+1=2`
`⇔(-4x²+4x²)+(8x+4x)+1=2`
`⇔12x+1=2`
`⇔12x=2-1`
`⇔12x=1`
`⇔x=1/12`
Vậy `x=1/12`
