a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

=>\(\left(x+2\right)^3=0\)

=>x+2=0

=>x=-2

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

b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)

c) đặt \(x^2+x+1=t\)

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

\(a,=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

Bài 4:

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

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

\(x^2-12x+36=\left(x-6\right)^2\)

\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)

\(=\left(x^2-x-6\right)^2-x^2=\left(x^2-6\right)\left(x^2-2x-6\right)\)

a) Ta có: \(4x^2-28xy+49y^2\)

\(=\left(2x\right)^2-2\cdot2x\cdot7y+\left(7y\right)^2\)

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

b) Ta có: \(x^2+8xy+16y^2\)

\(=x^2+2\cdot x\cdot4y+\left(4y\right)^2\)

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

c) Ta có: \(x^2-12x+36\)

\(=x^2-2\cdot x\cdot6+6^2\)

\(=\left(x-6\right)^2\)

\(\left(2x-7y\right)^2\)

\(\left(6-x\right)^2\)

4x² - 28xy + 49y²

= (2x)² - 2.2x.7y + (7y)²

= (2x - 7y)²

x² + 8x²y + 16x²y² đề có bị thiếu không , nên mình bổ sung nhé

= x² + 2.x.4xy + (4xy)²

= (x + 4xy)²

36 - 12x + x²

= 6² + 2.6.x + x²

= (6 + x)²

 Chúc bạn học tốt

a) 3x^4 - 12x^2 = 3x^2.(x^2 - 4) = 3x^2.(x - 2)(x + 2)

b) x^2 - 2xy + 3x - 6y

= x(x - 2y) + 3(x - 2y)

= (x - 2y)(x + 3) 

a) 3x^4 - 12x^2

= 3x^2.x^2- 3.4x^2

= x^2-4

b) x ^2 - 2xy + 3x - 6y

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

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

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

tự làm đi

\(a,=6y\left(2x^2-3xy-5y^2\right)\\ =6y\left(2x^2+2xy-5xy-5y^2\right)\\ =6y\left(x+y\right)\left(2x-5y\right)\\ b,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ c,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\\ =\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ d,=\left(a^2+3b\right)^2-1=\left(a^2+3b+1\right)\left(a^2+3b-1\right)\\ e,=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\\ =\left(2x-5\right)\left(2x+5-2x-7\right)\\ =-2\left(2x-5\right)\\ f,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ g,=x^3-x-6x-6\\ =x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ =\left(x+1\right)\left(x^2-3x+2x-6\right)\\ =\left(x+1\right)\left(x-3\right)\left(x+2\right)\\ l,=x^4+4x^2+4-4x^2\\ =\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\\ h,=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)