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

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

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

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

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

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

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

\(\left(x-5\right)\left(x-1\right)\left(x+3\right)\left(x+7\right)+60\)

\(=\left(x^2+2x-35\right)\left(x^2+2x-3\right)+60\)

\(=\left(x^2+2x\right)^2-38\left(x^2+2x\right)+105+60\)

\(=\left(x^2+2x\right)^2-3\left(x^2+2x\right)-35\left(x^2+2x\right)+165\)

\(=\left(x^2+2x-3\right)\left(x^2+2x-35\right)\)

\(=\left(x+3\right)\left(x-1\right)\left(x+7\right)\left(x-5\right)\)

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

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)

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

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

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

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

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

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

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

\(=(x+3)(x+5)(x-1)(x+1)\)

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

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

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

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

(x+3)(x+4)(x+5)(x+6)-24=[(x+3)(x+6)][(x+4)(x+5)]-24

                                  =(x²+6x+3x+3.6)(x²+5x+4x+5.4)-24

                                  =(x²+9x+18)(x²+9x+20)-24

                                  =(x²+9x+18)(x²+9x+18+2)-24    (*)

đặt x²+9x+18 là t   (1)

(*) trở thành

t(t+2)-24=t²+2t-24=t²-4t+6t-24

                          =(t²-4t)+(6t-24)

                          =t(t-4)+6(t-4)

                          =(t-4)(t+6)           (2)

thay (2) vào (1), ta được:

(x+3)(x+4)(x+5)(x+6)-24=(x²+9x+18-4)(x²+9x+18+6)

                                   =(x²+9x+14)(x²+9x+24)

                                   =(x²+7x+2x+14)(x²+9x+24)

                                   =[(x²+7x)+(2x+14)](x²+9x+24)

                                   =x(x+7)+2(x+7)(x²+9x+24)

                                   =(x+7)(x+2)(x²+9x+24)

(mình đã cố gắng giải thật chi tiết và phân tích triệt để nhất có thể rồi. có j sai sót thì góp ý nha!)

chả hiểu cái đề -_- 

( x + 3 )( x+ 4 )( x+  5 )( x+  6 ) - 24 

= ( x+ 3 )( x+ 6 )( x+ 4 )( x+ 5 ) - 24

 ( x^2 + 9x + 18 )( x^2 + 9x + 20 ) - 24 

Đặt x^2 + 9x + 19 = a 

= ( a - 1 )( a+ 1 ) - 24

= a^2 - 1 - 24 

= a^2 - 25 

= ( a- 5 )( a+ 5 ) 

= ( x^2 + 9x + 19 - 5 )( x^2 + 9x + 19 + 5 )

= ( x^2 + 9x + 14 )( x^2 + 9x + 24 ) 

x^5+x^4+1

=x⁵+x⁴+x³+x²+x+1-x³-x²-x

=x³.(x²+x+1)+(x²+x+1)-x.(x²+x+1)

tự xử tiếp

Minh Triều?????

Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)