Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)

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

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

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

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

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

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

(x+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?????

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

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)

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

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)

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

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

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~

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

\(=2x^2-3x+4x-6\)

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

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

Tham Khảo

2x² + x - 6

= 2x² + 4x - 3x - 6

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

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