c. x² - 5x + 4

= x² + x + 4x + 4

= x(x + 1) + 4(x + 1)

= (x + 4)(x + 1)

b. x² - 3x + xy - 3y

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

= (x + y)(x - 3)

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

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

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

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

1.

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

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

2.

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

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

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

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

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

\(x^2-7x+6=x^2-x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

a: =x(x-5)

a) \(=x\left(x-5\right)\)

b) \(=\left(x+3y-3y\right)\left(x+3y+3y\right)=x\left(x+6y\right)\)

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

Cách 1: Tách một hạng tử thành tổng hai hạng tử để xuất hiện nhân tử chung.

a) x2 – 3x + 2

= x2 – x – 2x + 2 (Tách –3x = – x – 2x)

= (x2 – x) – (2x – 2)

= x(x – 1) – 2(x – 1) (Có x – 1 là nhân tử chung)

= (x – 1)(x – 2)

Hoặc: x2 – 3x + 2

= x2 – 3x – 4 + 6 (Tách 2 = – 4 + 6)

= x2 – 4 – 3x + 6

= (x2 – 22) – 3(x – 2)

= (x – 2)(x + 2) – 3.(x – 2) (Xuất hiện nhân tử chung x – 2)

= (x – 2)(x + 2 – 3) = (x – 2)(x – 1)

b) x2 + x – 6

= x2 + 3x – 2x – 6 (Tách x = 3x – 2x)

= x(x + 3) – 2(x + 3) (có x + 3 là nhân tử chung)

= (x + 3)(x – 2)

c) x2 + 5x + 6 (Tách 5x = 2x + 3x)

= x2 + 2x + 3x + 6

= x(x + 2) + 3(x + 2) (Có x + 2 là nhân tử chung)

= (x + 2)(x + 3)

Cách 2: Đưa về hằng đẳng thức (1) hoặc (2)

a) x2 – 3x + 2

(Vì có x2 và  nên ta thêm bớt  để xuất hiện HĐT)

= (x – 2)(x – 1)

b) x2 + x - 6

= (x – 2)(x + 3).

c) x2 + 5x + 6

= (x + 2)(x + 3).

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

a) 3(x-y) - (x-y)^2

 =(x-y)(3-x+y)

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

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

-Đặt \(t=\left(x^2-x+1\right)\)

\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-4xt-xt+4x^2\)

\(=t\left(t-4x\right)-x\left(t-4x\right)\)

\(=\left(t-4x\right)\left(t-x\right)\)

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

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

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

CAM ON - HOANG

a: 5x-20xy

\(=5x\cdot1-5x\cdot4y=5x\left(1-4y\right)\)

b: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

c: \(x^2-2xy+y^2-z^2\)

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

=(x-y-z)(x-y+z)

d: \(5x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(5x-2\right)\)

e; \(x^2+4x+3=x^2+x+3x+3\)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f: \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

g: \(x^2-x-y^2-y\)

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

=(x-y)(x+y)-(x+y)

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

h: \(16x-5x^2-3\)

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

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

=(x-3)(-5x+1)

i: \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)

j: \(2x^2-6x=2x\cdot x-2x\cdot3=2x\left(x-3\right)\)

k: \(x^3-3x^2-4x+12\)

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

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

l: \(x^2-y^2-5x+5y\)

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

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