Gợi ý:

Nhóm:\(\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-8\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)

Đặt \(t=x^2+5x+4\) thì biểu thức trở thành:

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

Rồi bạn làm tiếp, nếu còn phân tích được thì phải phân tích, mình bận rồi.

(x + 1)(x + 2)(x + 3)(x + 4) - 8

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 8

= (x² + 4x + x + 4)(x² + 3x + 2x + 6) - 8

= (x² + 5x + 4)(x² + 5x + 6) - 8

Đặt x² + 5x + 5 = t

⇒ (x² + 5x + 5 - 1)(x² + 5x + 5 + 1) - 8 (1)

Thay t = x² + 5x + 5 vào (1), ta có:

(t - 1)(t + 1) - 8 = t² - 1 - 8 = t² - 9

= (t - 3)(t + 3)

⇔ (x² + 5x + 5 - 3)(x² + 5x + 5 + 3)

= (x² + 5x + 2)(x² + 5x + 8)

Chúc bạn học tốt !!!!!!!!

(x+1)(x+2)(x+3)(x+4)-8

= [(x+1)(x+4)][(x+2)(x+3)]-8

= (x²+4x+x+4)(x²+3x+2x+6)-8

= (x²+5x+5-1)(x²+5x+5+1)-8

= (x²+5x+5)²-1²-8

= (x²+5x+5)²-9

= (x²+5x+5) -3²

= (x²+5x+5-3)(x²+5x+5+3) {HĐT số 3}

= (x²+5x+2)(x²+5x+8)

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

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

c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)

d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)

a) \(x^2+4x+3\)

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

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

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

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

c) \(x^2-x-12\)

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

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

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

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

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

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

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

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

dat y^2+y=z cho gon

\(z^2-9z+20=z^2-4z-5z+20=z\left(z-4\right)-5\left(z-4\right)=\left(z-4\right)\left(z-5\right)\)

\(thaylai:\left(y^2+y-4\right)\left(y^2+y-5\right)\)

thay hay thi "k" ko hay thi thoi 

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

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

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

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

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

e)\(x^6-x^4-2x^3+2x^2=x^4\left(x^2-1\right)-2x^2\left(x-1\right)=x^4\left(x-1\right)\left(x+1\right)-2x^2\left(x-1\right)\)

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

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

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

a)\(x^2-x-12\)

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

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

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

b) \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

\(=x\left(x+3\right)+5\left(x+3\right)\)

\(=\left(x+3\right)\left(x+5\right)\)

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

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

2) \(=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left[\left(x+1\right)^2+\left(x+1\right).2y+4y^2\right]\)

\(=\left(x-2y+1\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)Đến đây bạn phân tích tiếp nha

CHÚC BẠN HỌC TỐT 

T I C K ủng hộ nha

\(x^4-4x^3-7x^2+22x+24\)

\(=\left(x^4-4x^3\right)-\left(7x^2-28x\right)-\left(6x-24\right)\)

\(=x^4.\left(x-4\right)-7x.\left(x-4\right)-6.\left(x-4\right)\)

\(=\left(x-4\right).\left(x^4-7x-6\right)\)

Tham khảo nhé~

đề có sai ko bn?

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

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

K sai dau
giao an truong Tran dai nghia do

x⁴+x³+2x²+x+1=x⁴+x³+x²+x²+x+1=(x⁴+x³+x²)+(x²+x+1)

                                                      =x²(x²+x+1)+(x²+x+1)

                                                       =(x²+x+1)(x²+1)

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

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

(x^+1)*(x^2+1+x0