`B1:`

`a)`

`x(x-2)+x-2=0`

`⇔(x-2)(x+1)=0`

`⇔` $\left[\begin{matrix} x-2=0\\ x+1=0\end{matrix}\right.$

`⇔` $\left[\begin{matrix} x=2\\ x=-1\end{matrix}\right.$

`b)`

`5x(x-3)-x+3=0`

`⇔5x(x-3)-(x-3)=0`

`⇔(x-3)(5x-1)=0`

`⇔` $\left[\begin{matrix} x-3=0\\ 5x-1=0\end{matrix}\right.$

`⇔` $\left[\begin{matrix} x=3\\ x=\dfrac{1}{5}\end{matrix}\right.$

`B2:`

`a)`

`x^3-2x^2+x`

`=x(x²-2x+1)`

`=x(x-1)²`

`b)`

` 2x^2+4x+2-2y^2`

`=2( x²+2x+1-y²)`

`=2[(x-1)²-y²]`

`=2(x-1-y)(x-1+y)`

`c)`

`2xy-x^2-y^2+16`

`=16-(x²-2xy+y²)`

`=4²-(x-y)²`

`=(4-x+y)(4+x-y)`

Hướng dẫn trả lời:

1)

a) `xcdot(x - 2) + x - 2 = 0`

`↔ xcdot(x - 2) + (x - 2) = 0`

`↔ xcdot(x - 2) + 1cdot(x - 2) = 0`

`↔ (x + 1)cdot(x - 2) = 0`

`↔ [(x + 1 = 0),(x - 2 = 0):}`

`↔ [(x = - 1),(x = 2):}`

Vậy `x = -1` hoặc `x = 2`

b) `5xcdot(x - 3) - x + 3 = 0`

`↔ 5xcdot(x - 3) - (x - 3) = 0`

`↔ 5xcdot(x - 3) - 1cdot(x - 3) = 0`

`↔ (5x - 1)cdot(x - 3) = 0`

`↔ [(5x - 1 = 0),(x - 3 = 0):}`

`↔ [(5x = 1),(x = 3):}`

`↔ [(x = 1/5),(x = 3):}`

Vậy `x = 1/5` hoặc `x= 3`

2)

a) `x^3 - 2x^2 + x`

`= xcdotx^2 - xcdot2x + xcdot1`

`= xcdot(x^2 - 2x + 1)`

`= xcdot(x^2 - 2cdotxcdot1 + 1^2)`

`= xcdot(x - 1)^2`

Giải thích:

Áp dụng HĐT `(A - B)^2 = A^2 - 2AB + B^2`

b) `2x^2 + 4x + 2 - 2y^2`

`= 2cdot(x^2 + 2x + 1 - y^2)`

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

`= 2cdot[(x^2 + 2cdotxcdot1 + 1^2) - y^2]`

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

`= 2cdot[(x + 1) + y]cdot[(x + 1) - y]`

`= 2cdot(x + y + 1)cdot(x - y + 1)`

Giải thích:

Áp dụng HĐT `(A + B)^2 = A^2 + 2AB + B^2` và HĐT `A^2 - B^2 = (A + B)cdot(A - B)`

c) `2xy - x^2 - y^2 + 16`

`= - (x^2 - 2xy + y^2 - 16)`

`= - [(x^2 - 2xy + y^2) - 16]`

`= - [(x - y)^2 - 4^2]`

`= - [(x - y) + 4]cdot[(x - y) - 4]`

`= - (x - y + 4)cdot(x - y - 4)`

Giải thích:

Áp dụng HĐT `(A - B)^2 = A^2 - 2AB + B^2` và HĐT `A^2 - B^2 = (A + B)cdot(A - B)`