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) \(2xy+6x+y^2+3y\)

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

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

x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung

a) \(4x^2+4x-3\)

\(=4x^2-2x+6x-3\)

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

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

b) \(2x^2+xy-y^2\)

\(=2x^2+2xy-xy-y^2\)

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

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

c) \(x^2+9x-22\)

\(=x^2+11x-2x-22\)

\(=x\left(x+11\right)-2\left(x+11\right)\)

\(=\left(x-2\right)\left(x+11\right)\)

a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)

b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

c) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)

d) \(y^2\left(x-1\right)-7y^3+7xy^3\)

\(=y^2\left(x-1-7y+7xy\right)\)

\(=y^2\left[\left(x-1\right)-7y\left(1-x\right)\right]=y^2\left(x-1\right)\left(1+7y\right)\)

a)

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

b)

 \(25-x^2+4xy-4y^2\\ =25-\left(x^2-4xy-4y^2\right)\\ =5^5-\left(x-y\right)^2\\ =\left(5+x-y\right)\left(5-x+y\right)\)

\(4x^2-y^2+8\left(y-2\right)=4x^2-y^2+8y-16\)

\(=4x^2-\left(y^2-8y+16\right)=4x^2-\left(y-4\right)^2\)

\(=\left(4x-y+4\right)\left(4x+y-4\right)\)

Bài 3. a) x(x-2)-2x+x=0

       <=> x²-2x-2x+x=0

       <=>x²-4x+x=0

       <=>x²-3x=0

      <=> x(x-3)=0 => x=0; x=3.

y^4+64

=(y^2)^2+16y^2+64-16y^2

=(y^2+8-4x)(x^2+8+4x)

x^2+4

=x^2+2x^2+4-2x^2

=(x+2)^2-2x^2

=(x^2+2-2x)(x^2+2+2x)

x^4+16

=(x^2)^2+4x^2+16-4x^2

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

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

x^4y^4+4

=x^4y^4+4x^4+2^2-4x^4

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

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

4x^4y^4+1

=4x^4y^4+x^4+1-x^4

=(2x^4y^4+1)^2-(x^2)^2

=(2x^4y^4+1-x^2)(2x^4y^4+1+x^2)

Mình ko bt câu D đúng hay sai nữa. Mà lỡ sai bạn đừng giận mình nha!

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

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

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

b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)

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

c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)