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4. Đặt  t= a^2 +a

Suy ra t^2 +4t - 12 = (t-2)(t+6) = (a^2+a-2) (a^2+a +6) = (a-1)(a+2)(a^2+a+6)

5. Đặt t = x^2 +x+1

Ta có: t(t+1) -12

= t^2 +t-12

= (t-3)(t+4)

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

 = (x-1) ( x+2) (x^2+x+5)

6. x^8 + x^7 + x^6 - x^7- x^6 - x^5 + x^5+ x^4 + x^3- x^4- x^3- x^2 + x^2 + x +1

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

7.  x^10 + x^9 +x^8 - x^9- x^8- x^7 +x^7+x^6+x^5 - x^6-x^5 - x^4 + x^5+ x^4 + x^3 - x^3 - x^2 - x + x^2 + x +1

=  (x^2 + x + 1) ( x^8 -x^7 + x^5 - x^4 + x^3 -x + 1)

         a3 - 7a - 6 

= a3 - a - 6a - 6 

= a ( a2 - 1 ) - 6 ( a + 1 )

= a ( a - 1 ) ( a + 1 ) - 6 ( a + 1 )

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

= ( a + 1 ) ( a2 - a - 6  )

= ( a + 1 ) ( a2 + 2a - 3a - 6 )

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

18 tháng 2 2021

 a) 3x2 – 7x + 2

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

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

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

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

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

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

.......?

 

 

 

 

a) Ta có: \(3x^2-7x+2\)

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

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

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

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

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

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

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

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

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

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

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

20 tháng 7 2021

a, \(4abc-8ab^2c=4abc\left(1-2b\right)\)

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

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

c, \(9a^4\left(a-2\right)+a^2\left(a-2\right)=a^2\left(9a^2+1\right)\left(a-2\right)\)

d, \(\left(a-4\right)\left(2a-1\right)-8a+4=\left(a-4\right)\left(2a-1\right)-4\left(2a-1\right)\)

\(=\left(a-8\right)\left(2a-1\right)\)

20 tháng 7 2021

a) `4abc-8ab^2c=4abc(1-2b)`

b) `x^2 (2a-1)+x(1-2a) = x^2 (2a-1) -x(2a-1) = (2a-1)(x^2-x)=x(2a-1)(x-1)`

c) `9a^4 (a-2) +a^2 (a-2) = (a-2)(9a^4+a^2)=a^2 (a-2)(9a^2+1)`

d) `(a-4)(2a-1)-8a+4=(a-4)(2a-1)-4(2a-1)=(2a-1)(a-8)`

8 tháng 11 2017

2\

a3+4a2-7a-10

= a3-2a2+6a2-12a+5a-10

=a2(a-2) +6a(a-2) +5(a-2)

= (a-2)(a2+6a+5)

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

4\

(a2+a)2+4(a2+a)-12

= (a2+a)2+4(a2+a)+4-16

= (a2+a+2)2-16

= (a2+a+6)(a2+a-2)

5/

(x2+x+1)(x2+x+2)-12

đặt x2+x+1=a

⇒ a(a+1)-12

= a2+a-12

= a2-3a+4a-12

= a(a-3)+4(a-3)

= (a-3)(a+4)

⇒ (x2+x-2)(x2+x+5)

6\

x8+x+1

= x8+x7+x6-x7-x6-x5+x5+x4+x3-x4-x3-x2+x2+x+1

= x6(x2+x+1) - x5(x2+x+1) +x3(x2+x+1)-x2(x2+x+1)+(x2+x+1)

= (x2+x+1)(x6-x5+x3+x2+1)

7\

x10+x5+1

= x10+x9+x8-x9-x8-x7+x7+x6+x5-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1

= x8(x2+x+1)-x7(x2+x+1)+x5(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)

= (x2+x+1)(x8-x7+x5-x4+x3-x+1)

AH
Akai Haruma
Giáo viên
15 tháng 5 2021

1.

$a^3-7a-6=a^3-a-(6a+6)=a(a^2-1)-6(a+1)$

$=a(a-1)(a+1)-6(a+1)=(a+1)(a^2-a-6)$

$=(a+1)(a^2+2a-3a-6)$

$=(a+1)[a(a+2)-3(a+2)]=(a+1)(a+2)(a-3)$

2.

\(a^3+4a^2-7a-10=a^3+a^2+(3a^2+3a)-(10a+10)\)

\(=a^2(a+1)+3a(a+1)-10(a+1)=(a+1)(a^2+3a-10)\)

\(=(a+1)[a(a-2)+5(a-2)]=(a+1)(a-2)(a+5)\)

AH
Akai Haruma
Giáo viên
15 tháng 5 2021

3.

\(a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc\)

\(=a(b^2+c^2+2bc)+b(c^2+a^2+2ac)+c(a^2+b^2+2ab)-4abc\)

\(=ab(a+b)+bc(b+c)+ca(c+a)+2abc\)

\(=ab(a+b+c)+bc(a+b+c)+ac(a+c)\)

\(=(a+b+c)(ab+bc)+ac(a+c)=(ab+b^2+bc)(a+c)+ac(a+c)\)

\(=(a+c)(ab+b^2+bc+ac)=(a+c)(a+b)(b+c)\)

 

27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

17 tháng 7 2021

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

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

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

 

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

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