\(\left(x-2\right)\left(x-1\right)x\left(x+1\right)-24\)

\(=\left(x^2-x-2\right)\left(x^2-x\right)-24\)

\(=\left(x^2-x\right)-2\left(x^2-x\right)-24\)

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

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

đây là cách mình chế ra bạn ko hiểu chỗ nào hỏi mk nhé

\(3x^2-14x-5=3x\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(3x+1\right)\)

\(3x^2-14x-5\)

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

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

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

A= 3x^2 + 14x - 15
3A= (3x)^2 + 2.7.(3x) - 15.3
3A= (3x+7)^2 +(15.3-7.7)
3A= (3x+7 )^2 -2^2
3A= (3x+7-2 )((3x+7+2 )
A= ( 3x-5)( x+ 3)

\(9x^4+36x^3+29x^2-14x\)

\(=x\left(9x^3+36x^2+29x-14\right)\)

\(=x\left(9x^3+18x^2+18x^2+36x-7x-14\right)\)

\(=x\left[9x^2\left(x+2\right)+18x\left(x+2\right)-7\left(x+2\right)\right]\)

\(=x\left(x+2\right)\left(9x^2+18x-7\right)\)

\(=x\left(x+2\right)\left(9x^2-3x+21x-7\right)\)

\(=x\left(x+2\right)\left[3x\left(3x-1\right)+7\left(3x-1\right)\right]\)

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

(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 ) 

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

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

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

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

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

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

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

\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)

\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)