1.Phân tích thành nhân tử ( phương pháp nhóm nhiều hạng tử )

a. x^3 + 2x^2 - xy - 2y

\(=x^2\left(x+2\right)-y\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-y\right)\)

b. xy - 5x + 3y^2 - 15y

\(=xy+3y^2-5x-15y\)

\(=y\left(x+3y\right)-5\left(x+3y\right)\)

\(=\left(x+3y\right)\left(y-5\right)\)

c.2xy + 6x + y^2 + 3y

\(=2xy+y^2+6x+3y\)

\(=y\left(2x+y\right)+3\left(2x+y\right)\)

\(=\left(2x+y\right)\left(y+3\right)\)

a) \(x^3+2x^2-xy-2y\)

\(=\left(x^3-xy\right)+\left(2x^2-2y\right)\)

\(=x\left(x^2-y\right)+2\left(x^2-y\right)\)

\(=\left(x+2\right)\left(x^2-y\right)\)

\(=\left(x+2\right)\left(x+\sqrt{y}\right)\left(x-\sqrt{y}\right)\)

a) \(x^3+2x^2-xy-2y\)

\(=x^2\left(x+2\right)-y\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-y\right)\)

\(=\left(x+2\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

b) \(xy-5x+3y^2-15y\)

\(=x\left(y-5\right)+3y\left(y-5\right)\)

\(=\left(y-5\right)\left(x+3\right)\)

c) \(2xy+6x+y^2+3y\)

\(=2x\left(y+3\right)+y\left(y+3\right)\)

\(=\left(y+3\right)\left(2x+y\right)\)

a) 5x^2 + 6xy + y^2

=5x²+5xy+xy+y²

=5x.(x+y)+y.(x+y)

=(x+y)(5x+y)

b) x^2 + 2xy - 15y^2.

=x²-3xy+5xy-15y²

=x.(x-3y)+5y.(x-3y)

=(x-3y)(x+5y)

c) (x-y)^2 + 4(x-y) - 12

=(x-y)²+4(x-y)+4-16

=(x-y+2)²-16

=(x-y+2-4)(x-y+2+4)

=(x-y-2)(x-y+6)

d) x^3 - 2x - 4.

=x³+2x²+2x-2x²-4x-4

=x.(x²+2x+2)-2.(x²+2x+2)

=(x²+2x+2)(x-2)

a: =5(2x+3y)

d: =(x+1-y)(x+1+y)

`1)`

`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`

`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`

`2)`

`a)2x(x-3)-x+3=0`

`<=>2x(x-3)-(x-3)=0`

`<=>(x-3)(2x-1)=0`

`<=>[(x=3),(x=1/2):}`

`b)x^2+5x+6=0`

`<=>x^2+2x+3x+6=0`

`<=>(x+2)(x+3)=0`

`<=>[(x=-2),(x=-3):}`

a) Biến đổi x 3   =   x 2 .x, phân tích thành x( x 2  + 2).

b) Tương tự a) phân tích thành 3(x – 2y).

c) Nhân tử chung 5(x + 3y) phân tích thành 5(x + 3y)(1 – 3x).

d) Thực hiện biến đổi y – x = -(x – y), xuất hiện nhân tử chung là (x – y), phân tích thành (x – y)(3 + 5x).

a) \(2x\left(x-7\right)-5y\left(x-7\right)=\left(x-7\right)\left(2x-5y\right)\)

b) \(5x^3y+10x^2y+5xy=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(4y^2-4y-x^2+1=\left(2y-1\right)^2-x^2=\left(2y-1-x\right)\left(2y-1+x\right)\)

d) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)

a: \(=\left(x-7\right)\left(2x-5y\right)\)

b: \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

a) \(=\left(2x-5y\right)\left(x-7\right)\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(=\left(\left(4y^2-4y+1\right)-x^2\right)=\left(2y-1\right)^2-x^2=\left(2y-1+x\right)\left(2y-1-x\right)\)