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

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

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

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

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

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

\(b,\)\(3x-7x-10\)

\(=3x^2+3x-10x-10\)

\(=\left(3x^2+3x\right)-\left(10x+10\right)\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(3x-10\right)\left(x+1\right)\)

\(c,\)\(x^4+1-2x^2\)

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

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

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

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

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

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

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

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

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

B1

A=11x^2-x-2

B=2(-4+x)

B2

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

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

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

2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)

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

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

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

5: \(y^2\left(x^2+y\right)-z\cdot x^2-zy\)

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

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

7: \(5\left(x+y\right)^2+15\left(x+y\right)\)

\(=5\left(x+y\right)\cdot\left(x+y\right)+5\left(x+y\right)\cdot3\)

=5(x+y)(x+y+3)

9: \(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)

\(=\left(y-4\right)^2\left(7x+y-4\right)\)

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

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

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

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

2: \(5x\left(x-2\right)-3x^2\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-3x^2\right)=x\left(5-3x\right)\left(x-2\right)\)

4: 3x(x-5y)-2y(5y-x)

=3x(x-5y)+2y(x-5y)

=(x-5y)(3x+2y)

6: b(a-c)+5c-5a

=b(a-c)-5(a-c)

=(a-c)(b-5)

8: 9x(x-y)-10(y-x)^2

=9x(x-y)-10(x-y)^2

=(x-y)(9x-10x+10y)

=(x-y)(-x+10y)

10: \(\left(a-b\right)^2-\left(a+b\right)\left(b-a\right)\)

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

=(a-b)(a-b+a+b)

=2a(a-b)

12: 2x(x-3)+y(x-3)+(3-x)

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

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

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

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

\(H=-x^2+5x-4=-x^2+4x+x-4\)

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

\(I=x^2+4x+3=x^2+3x+x+3\)

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

\(K=2x^2+7x+5=2x^2+2x+5x+5\)

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

\(L=-3x^2-5x-2=-3x^2-3x-2x-2\)

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

G = 2x² - 3x +1 =  2x² -2x -x +1 =(x-1).(2x-1)

H = -x² + 5x - 4 = -x² + 4x +x-4 = (x-4).(1-x)

I = x² + 4x + 3 = x² + 3x + x + 3 =(x+3).(x+1)

K = 2x² + 7x + 5 = 2x² + 2x + 5x + 5 = (x+1).(2x+5)

L = -3x² -5x -2 = -3x² - 3x - 2x - 2 = -3.x(x+1) - 2.(x+1) = (x+1).(-3x-2)

a)3x^2-7x+2

=3x^2-x-6x+2

=(3x^2-x)-(6x-2)

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

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

cho loi nx nha may ban

Mình mới lớp 7 nên chỉ giải được 1 bài thôi!

\(3x^2-7x+2=3x^2-\left(6x+1x\right)+2=3x^2-6x-1x+2\)

\(3x\left(x-2\right)-1\left(x-2\right)=\left(x-2\right)\left(3x-1\right)=3\left(x-2\right)\left(x-\frac{1}{3}\right)\)

Ở đầu dòng cuối cùng bạn thêm dấu "=" giúp mình nhé! Nãy mình đánh thiếu!