1)   \(3x^2+2x-1\)

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

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

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

2)   \(x^3+6x^2+11x+6\)

\(=x^3+x^2+5x^2+5x+6x+6\)

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

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

\(=\left(x+1\right)\left(x+2x+3x+6\right)\)

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

3)   \(x^4+2x^2-3\)

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

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

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

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

4)   \(ab+ac+b^2+2bc+c^2\)

\(=a\left(b+c\right)+\left(b+c\right)^2\)

\(=\left(b+c\right)\left(a+b+c\right)\)

1, \(3x^2+2x-1\)

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

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

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

2, \(x^3+6x^2+11x+6\)

\(=\left(x^3+3x^2\right)+\left(3x^2+9x\right)+\left(2x+6\right)\)

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

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

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

a: =4x^2+8x-3x-6

=4x(x+2)-3(x+2)

=(x+2)(4x-3)

b: =3(3x^2-2x-1)

=3(3x^2-3x+x-1)

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

c: =2x^2-4x+x-2

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

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

d: =3x^2+3x-2x-2

=3x(x+1)-2(x+1)

=(x+1)(3x-2)

e: =3x^2+9x+x+3

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

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

a) \(4x^2+5x-6\)

\(=4x^2+8x-3x-6\)

\(=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

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

b) \(9x^2-6x-3\)

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

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

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

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

c) \(2x^2-3x-2\)

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

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

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

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

d) \(3x^2+x-2\)

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

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

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

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

e) \(3x^2+10x+3\)

\(=3x^2+9x+x+3\)

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

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

đề dài nên T giải câu a thôi bn tự làm tiếp mấy câu khác nhé 

2x^2 - 2y^2 - 6x - 6y 

= 2(x^2-y^2) - 6(x+ y) 

= 2(x-y)(x+y) - 6(x+y) 

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

ko biết

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

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

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

Ta có 

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

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

2, \(x^3+2x^2+4x^2+8x+3x+6\)

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

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

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

\(=\left(x+2\right)\text{[}x\left(x+1\right)+3\left(x+1\right)\text{]}\)

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

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

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

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

4,\(ab+ac+b^2+2bc+c^2\)

\(=a\left(b+c\right)+\left(b+c\right)^2\)

\(=\left(b+c\right)\left(a+b+c\right)\)

 3x^2+2x-1 
=3x^2+3x-x-1 
=3x(x+1)-(x+1) 
=(x+1)(3x-1) 

x^3+6x^2+11x+6 
=x^3+5x^2+6x+x^2+5x+6 
=x(x^2+5x+6)+(x^2+5x+6) 
=(x+1)(x^2+5x+6) 
=(x+1)(x^2+3x+2x+6) 
=(x+1)(x+2)(x+3) 

x^4+2x^2-3 
=x^4-x^2+3x^2-3 
=x^2(x^2-1)+3(x^2-1) 
=(x^2-1)(x^2+3) 
=(x+1)(x-1)(x^2+3) 

ab+ac+b^2+2bc+c^2 
=a(b+c)+(b+c)^2 
=(b+c)(a+b+c) 

a^3-b^3+c^3+3abc 
=(a-b)^3+3ab(a-b)+c^3+3abc 
=(a-b+c)^3-3(a-b)c(a-b+c)+3ab(a-b+c) 
=(a-b+c)(a^2+b^2+c^2-2ab+2ac-2bc-3ac+3... 
=(a-b+c)(a^2+b^2+c^2+ab+bc-ca) 
=1/2.(a-b+c)(a^2+2ab+b^2+b^2+2bc+c^2+c... 
=1/2.(a-b+c)[(a+b)^2+(b+c)^2+(c-a)^2]

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^