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]

3,  \(=x^4-x^2+3x^2-3\)

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

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

5,  nhận xét  : \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\Rightarrow a^3-b^3=\left(a-b\right)^3+3a^2b-3ab^2\)

thay vào đầu bài ta có:  \(\left(a-b\right)^3+c^3+3a^2b-3ab^2+3abc\)

\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)

\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-ac+bc\right)\)

1, \(=3x^2+3x-x-1\)

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

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

3x² + 2x – 1

=3x² -x +3x-1

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

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

sắp nôn       

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

2)\(x^3+6x^2+11x+6=x^3+3x^2+3x^2+9x+2x+6\)

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

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

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

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

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

4)\(ab+ac+b^2+2bc+c^2=a\left(b+c\right)+\left(b+c\right)^2=\left(b+c\right)\left(a+b+c\right)\)

5) câu này sau khi phân tích được (a-b+c)(a²+b²+c²+ab+bc-ac)

\(a,3x^2+2x-1\)

\(\Leftrightarrow3x^2+3x-x-1\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)\)

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

\(b,x^3+6x^2+11x+6\)

\(\Leftrightarrow x^3+3x^2+3x^2+9x+2x+6\)

\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+6\right)\)

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

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

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

\(\Leftrightarrow x^4-x^3+x^3-x^2+3x^2-3\)

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

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

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

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

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

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

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

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

a) \(3x^2+2x-1=3x^2+3x-x\)\(-1\)

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

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

b) x³ + 6x² + 11x + 6 = x³+6x²+12x+8-x-2 = ( x³+6x²+12x+8)-(x+2) = (x+2)^3 - (x+2) = (x+2)[(x+2)^2 - 1]

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

c) x⁴ + 2x² – 3 = x⁴ + 2x² +1 - 4 = (x^2 +1)^2 - 2^2

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

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

​= a(b+c) + (b+c)^2 

=(b+c)(a+b+c)

\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2+c^3+3abc\)

\(=\left(a-b\right)^3+3ab\left(a-b\right)+c^3+3abc\)

\(=\left[\left(a-b\right)^3+c^3\right]+\left[3ab\left(a-b\right)+3abc\right]\)

\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]\)\(+3ab\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2\right)\)\(+3ab\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)

\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)

câu cuối ko bik đúng hay sai đâu nha

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

b thiếu đề bài nè  x^4 - 2x^2 - 3 = 0 
(x^2)(x^2)-2(x^2)-3=0 
(x^2)(x^2)-3(x^2)+1(x^2)-3=0 
(x^2)(x^2-3) + 1(x^2-3) = 0 
(x^2-3) (x^2+1)=0 

c bó tay

d (a3 + b3 + c3) - 3abc
= ( (a+b+c)3 - 3ab(a+b) -3bc(b+c) -3ac(a+c) - 6abc) - 3abc 
= (a+b+c)3 - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 9abc 
= (a+b+c)3 
- 3ab(a+b) - 3abc 
- 3bc(b+c) - 3abc 
- 3ac(a+c) - 3abc 
= (a+b+c)3 
- 3ab(a+b+c) 
- 3bc(a+b+c) 
- 3ac(a+b+c) 
= (a+b+c)( (a+b+c)2 - 3ab -3bc 3ac) 
=(a+b+c)( a2 + b2 + c2 + 2ab +2bc + 2ca -3ab - 3bc -3ac) 
=(a+b+c) (a2 + b2 + c2 - ab - bc -ac) 

ý d hình như đề sai

chị ui kết bạn với em đi em hết lượt kết bạn rùi 

em học lớp 5 

1. x³ + 6x² + 11x + 6

= x³ + 3x² + 3x² + 9x + 2x + 6

= x²(x + 3) + 3x(x + 3) + 2(x + 3)

= (x + 3)(x² + 3x + 2)