Ta có : x³ - 7x + 6 

= x³ - x - 6x + 6 

= x(x² - 1) - 6(x - 1)

= x(x + 1)(x - 1) - 6(x - 1)

= (x - 1) [x(x + 1) - 6]

= (x - 1) (x² + x - 6) . 

CÁC Ý SAU TƯƠNG TỰ

   x³ - 7x + 6 

= x³ - x - 6x + 6 

= x(x² - 1) - 6(x - 1)

= x(x + 1)(x - 1) - 6(x - 1)

= (x - 1) [x(x + 1) - 6]

= (x - 1) (x² + x - 6) . 

1) Ta có : 2x² + 3x - 5

= 2x² - 2x + 5x - 5

= 2x(x - 1) + 5(x - 1)

= (x - 1) (2x + 5) 

3) x² + x - 6

= x² + 2x - 3x - 6

= x(x + 2) - (3x + 6)

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

= (x - 3)(x + 2) 

1

x³-7x+6

=x³+0x²-7x +6

= x³-x²+x²-x-6x+6

=(x³-x²)+(x²-x)-(6x-6)

=x²(x-1)+x(x-1)-6(x-1)

=(x-1)(x²+x-6)

=(x-1)(x²+3x-2x-6)

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

=(x-1)(x-2)(x+3)

7) (x+2)(x+3)(x+4)(x+5)-24

=(x+2)(x+5) (x+3)(x+4)-24

=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24

=[x²+5x+2x+10][x²+4x+3x+12]-24

=[x²+7x+10][x²+7x+12]-24

đặt a=x²+7x+10

=>x²+7x+12=a+2

=a(a+2)-24

=a²+2a-24

=a²+6a-4a-24

=(a²+6a)-(4a+24)

=a(a+6)-4(a+6)

=(a+6)(a-4)

thay a= x²+7x+10 vào ta được

(x²+7x+10+6)(x²+7x+10-4)

=(x²+7x+16)(x²+7x+6)

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

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

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

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

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

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

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

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

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

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

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

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

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

Mình sửa: Bài 1
2)x²+3x-15

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

\(x^2-7x+10\)

\(=x^2-2x-5x+10\)

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

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

học tốt

\(x^3-x^2-8x+12\)

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

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

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

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