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 )

1. a³ - 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 -3a + 2a - 6)

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

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

2. a³ + 4a² - 7a - 10

= a^3 + 5a^2 - a^2 - 5a - 2a - 10

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

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

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

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

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

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

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

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

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

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

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

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

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

\(a,a^3-7a-6\)

\(\Leftrightarrow a^3+a^2-a^2-a-6a-6\)

\(\Leftrightarrow a^2\left(a+1\right)-a\left(a+1\right)-6\left(a+1\right)\)

\(\Leftrightarrow\left(a+1\right)\left(a^2-a-6\right)\)

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

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

\(\Leftrightarrow a^3+5a^2-a^2-5a-2a-10\)

\(\Leftrightarrow a^2\left(a+5\right)-a\left(a+5\right)-2\left(a+5\right)\)

\(\Leftrightarrow\left(a+5\right)\left(a+1\right)\left(a-2\right)\)

\(d,\left(a^2+a\right)^2+4\left(a^2+a\right)-12\)

Đặt a^2+a=y ta có 

y^2+4y-12=(y+6)(y-2)

<=> (a^2+a+6)(a^2+a-2)

<=> (a^2+a+6)(x-1)(x+2)

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

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

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

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)`

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

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: \(x^4+2x^3-4x-4\)

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

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

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

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

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

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

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)