Ta có :

f(x) = x³-2x²+5x+a = x³ - 3x² + x² -3x + 8x + a

      = x²(x-3) + x(x-3) + 8x + a

Vì x²(x-3) + x(x-3) ⋮ x-3 

=> 8x + a có dạng (x-3)k = kx - 3k

Khi đó k = 8 và a có dạng -3k = -3 . 8 = -24

để f(x) = x³ - 2x² +5x + a  ⋮ x - 3

thì số dư của phép chia f(x) cho x - 3 bằng 0

áp dụng  định lí BÉZOUT ta có 

f(3) = 3³ - 2.3² +5.3 + a = 0⇔ 27-18+15 + a =0

⇔ 9+ 15 + a = 0 ⇔ a = -24

2010x² - x -2011 = 0

=> 2011x² - 2011 - x²-x = 0

=> 2011(x²-1) - x(x+1) =0

=> 2011(x-1)(x+1) - x(x+1) = 0

=> (x+1)[2011(x-1)-x]=0

=> (x+1)(2011x-x-2011)=0

=> (x+1)(2010x-2011)=0

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2010x-2011=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\2010x=2011\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2011}{2010}\end{matrix}\right.\)

`a^2b-a^2+ab-a^3`

`=(a^2b-a^3)+(-a^2+ab)`

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

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

`=a(b-a)(a+1)`

   a²b - a² + ab - a³ 

= (a²b +ab) - a²(a+1)

= ab(a+1) - a²(a+1)

=(a+1)(ab-a²)

=a(b-a)(a+1)

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

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

\(=\left(b-a\right)a\left(a+1\right)\)

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

8-(x-1)3 = 23-(x-1)3

= (2-x+1)[4+2(x-1)+(x-1)2]

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

=(3-x)(4+4x)=4(3-x)(1+x)

làm như nào đấy chỉ vs :vv

a)x² - 4x -5=x² + x - 5x -5=x(x+1)-5(x+1)=(x+1)(x-5)

b)x² + 5x - 14=x² +7x -2x -14=x(x-2)+7(x-2)=(x-2)(x+7)

c)3x² - 5x +2 =3x²-3x-2x+2=3x(x-1)-2(x-1)=(x-1)(3x-2)

x² - 4x - 5 = x² - 4x + 4 - 9 = (x-2)² - 9 =(x-5)(x+1)

x² + 5x - 14 = x² - 2x  + 7x - 14 = x(x-2) +7(x-2) = (x-2)(x+7)

3x² - 5x + 2 = 3x² -3 -5x + 5 = 3(x-1)(x+1) -5(x-1)=(x-1)(3x-2)

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

= 2x² + 6x + 2x² +3x - 3x - 2

= 4x² + 6x - 2

= (2x)² + 2.2x.\(\dfrac{3}{2}\)+\(\dfrac{9}{4}-\dfrac{17}{4}\)

= \(\left(2x-\dfrac{3}{2}\right)^2-\dfrac{17}{4}\)

b) (x+2)(x²-2x+4) + (x+2)(x²+2x+1)

= (x+2)[(x²-2x+4)+(x²+2x+1)]

= (x+2)(2x²+5)

= 2x³+5x+4x²+10

= 2x³ + 4x² + 5x + 10

= 2x²(x+2) + 5(x+2)

= ( 2x² + 5 )(x+2)

c) (x-3)(x+3) - 2(x-1)(x+2) + (x+5)(x-2)

= x²-9-(2x-2)(x+2)+(x²+3x-10)

= x²-9-2x²+2x-4+x²+3x-10

= 5x - 23

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

= 2x2 + 6x + 2x2 +3x - 3x - 2

= 4x2 + 6x - 2

= (2x)2 + 2.2x.$\genfrac{}{}{0ex}{}{3}{2}$+$\genfrac{}{}{0ex}{}{9}{4}-\genfrac{}{}{0ex}{}{17}{4}$

= ${\left(2x-\genfrac{}{}{0ex}{}{3}{2}\right)}^2-\genfrac{}{}{0ex}{}{17}{4}$

b) (x+2)(x2-2x+4) + (x+2)(x2+2x+1)

= (x+2)[(x2-2x+4)+(x2+2x+1)]

= (x+2)(2x2+5)

= 2x3+5x+4x2+10

= 2x3 + 4x2 + 5x + 10

= 2x2(x+2) + 5(x+2)

= ( 2x2 + 5 )(x+2)

c) (x-3)(x+3) - 2(x-1)(x+2) + (x+5)(x-2)

= x2-9-(2x-2)(x+2)+(x2+3x-10)

= x2-9-2x2+2x-4+x2+3x-10

= 5x - 23