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

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

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)(1)

Đặt \(x^2+5x=t\)

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

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

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

Mà \(x^2+5x=t\)nên \(\left(2\right)=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)

hay \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)\(=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)

x^7+x+1

=x.x^6+x.1+x.1/x

=x.(x^6+1+1/x)

tk 

Sửa đề x^7 chuyển thành x^8

Ta có

\(x^8+x+1=x^8-x^2+x^2+x+1\)

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

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

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

\(=\left(x^2+x+1\right)\left(x^6+x^3-x^5-x^2+1\right)\)

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

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

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

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

Học tốt nhé

x^2 - 3x + 2

= x^2 - 2x - x + 2

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

= (x-2)(x-1)

hok tốt

Sai đề bài rồi bạn ơi

(4x+1)(12x-1)(3x+2)(x+1)=4 
<=> [(4x+1)(3x+2)].[(12x-1)(x+1)]=4 
<=>(12x^2+11x+2)(12x^2+11x-1)=4 
Đặt 12x^2+11x+2=t thì 12x^2+11x-1=t-3, thay vào phương trình trên ta có: 
pt<=>t(t-3)=4 
<=> t^2-3t-4=0 
<=> (t-4)(t+1)=0 
<=> t=4 hoặc t=-1 
Thay t=12x^2+11x+2, có: 
12x^2+11x+2=4 (1) hoặc 12t^2+11x+2= -1 (2) 
Giải pt(1), ta có nghiệm x= [-11+ (căn bậc hai của (217)]/24 hoặc x= [-11-(căn bậc hai của(217)]/24 
giải pt(2), ta thấy vô nghiệm.

( 4x + 1 ) ( 12x - 1 ) ( 3x + 2 ) ( x + 1 ) - 4

= ( 12x² + 11x - 1 ) ( 12x² + 11x + 2 ) - 4

Đặt 12x² + 11x - 1 = a , ta có :

y² + 3y - 4 = ( y - 1 ) ( y + 4 )

                 = ( 12x² + 11x - 2 ) ( 12x² + 11x + 6 ) 

..... 

ko chắc

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

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

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

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

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

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

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

phân tích kiểu j bn??? xem lại đề bài ik nha

x+2=2.\(\frac{1}{2}\).x+2=2.(\(\frac{x}{2}\)+1)

a, x²-5xy+2x-10y = (x² + 2x)-(5xy+10y)

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

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

b, x²-5x+4 = x²- x - 4x +4

                  = (x²-x)-(4x-4)

                  =x(x-1)-4(x-4)

                  =(x-1)(x-4)

\(a,x^2-5xy+2x-10y\)

\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)

\(=x\left(x-5y\right)+2\left(x-5y\right)\)

\(=\left(x-5y\right)\left(x+2\right)\)

\(b,x^2-5x+4\)

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

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

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

\(x^6+x^4+x^2y^2+y^4-y^6\)

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

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

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

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