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

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

\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)

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

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

\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)

A = x² + 3x + 2

A = x² + x + 2x + 2

A = x(x + 1) + 2(x + 1)

A = (x + 2)(x + 1)

Lời giải:
a.

$x^2-7x+6=(x^2-x)-(6x-6)=x(x-1)-6(x-1)=(x-1)(x-6)$

b.

$x-3\sqrt{3}x-12\sqrt{3}$ không phân tích được thành nhân tử

c.

$x^2+4x-2$ không phân tích được thành nhân tử với các hệ số nguyên.

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

b) \(3x^2+4x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(3x+7\right)\)

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

a) (x+1)(x+4)

b) (x-1)(3x+7)

c) (x+3)(x+4)

a) x²+5x+4 = x(x+4)+(x+4) = (x+4)(x+1)

b) 3x²+4x-7 = 3x(x-1)+7(x-1) = (x-1)(3x+7)

c) x²+7x+12 = x(x+4)+3(x+4) = (x+3)(x+4)

Phân tích đa thức thành nhân tử:

a) x² (2x - 5) + 6x - 15 =  x² (2x - 5) + 3(2x - 5) =( x2 + 3 )(2x - 5)

b)x² + 7x + 12 =  x² + 3x + 4x + 12 = (x + 3 )( x + 4)

a: \(7x-14y=7\left(x-2y\right)\)

b: \(4x^2-4x+1=\left(2x-1\right)^2\)

b) x²-3x+xy-3y

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

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

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

c) x²-y²-4x+4

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

=\(\left(x-2\right)^2\) \(-y^2\)

=(\(x-y-2\)) \(\left(x+y-2\right)\)

a)6x³y-9x²y²+3xy

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

a) x³+4x-5 = x³-x²+x²+4x-5=(x³-x²)+(x²-x)+(5x-5)=x²(x-1)+x(x-1)+5(x-1)=(x²+x+5)(x-1)

b) x³-3x²+4=x³-2x²-x²+4=(x³-2x²)-(x²-4)=x²(x-2)-(x-2)(x+2)=(x²-x+2)(x-2)

c) x³+2x²+3x+2=x³+x²+x²+x+2x+2=(x³+x²)+(x²+x)+(2x+2)=x²(x+1)+x(x+1)+2(x+1)=(x²+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x²+3x)²-2(x²+3x)-8=x⁴+6x³+9x²-2x²-6x-8=x⁴+6x³+7x²-6x-8=x⁴-x³+7x³-7x²+14x²-14x+8x-8=(x⁴-x³)+(7x³-7x²)+(14x²-14x)+(8x-8)=x³(x-1)+7x²(x-1)+14x(x-1)+8(x-1)=(x³+7x²+14x+8)(x-1)=(x³+x²+6x²+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x²+4x+10)²-7(x²+4x+11)+7=(x²+4x+10)²-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

a) Ta có: \(x^3+4x-5\)

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

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

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

b) Ta có: \(x^3-3x^2+4\)

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

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

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

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

c) Ta có: \(x^3+2x^2+3x+2\)

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

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

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

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

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

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

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

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

e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)

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

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

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

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

f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)

\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)

\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)

\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)

\(a,\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (sửa \(2x\rightarrow2x^2\)

Đặt \(x^2+4x+8=a\)

\(=a^2+3ax+2x=a^2+ax+2ax+2x^2=\left(a+x\right)\left(a+2x\right)\\ =\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)

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

\(=\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)

\(=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

a: =3x^2-3x-8x+8=(x-1)(3x-8)

b: =x^2-x-5x+5=(x-1)(x-5)

c: =x^2-6x+2x-12=(x-6)(x+2)

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

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

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

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