Ta có (x³ + ax² + bx + 3) : (x² - 2x - 1) = x + a - 2 dư x(b - 2a + 5) + a + 1

Để  (x³ + ax² + bx + 3) \(⋮\) (x² - 2x - 1)

=> x(b - 2a + 5) + a + 1 = 0 \(\forall x\)

=> \(\hept{\begin{cases}b-2a+5=0\\a+1=0\end{cases}}\Rightarrow\hept{\begin{cases}b-2a=-5\\a=-1\end{cases}}\Rightarrow\hept{\begin{cases}b=-7\\a=-1\end{cases}}\)

Bai lam 

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

=3x(x-2y)

a) x³ - 2x² + x 

= x(x² - 2x + 1)

= x(x - 1)²

b) x² - 2x - 15

= x² - 2x + 1 - 16

= (x - 1)² - 4²

= (x - 5)(x + 3)

c) 5x²y³ - 25x³y⁴ + 10x³y³

= 5x²y³(1 - 5xy + 2x)

d) 12x²y - 18xy² - 30y²

= 6y(2x² - 3xy - 5y) 

= 6y(2x² + 2xy - 5xy - 5y)

= 6y[2x(x + y) - 5y(x + y)

= 6y(x + y)(2x - 5y)

e) 5(x - y) - y(x - y)

= (5 - y)(x - y)

g) 36 - 12x + x²

= (6 - x)²

h) 4x² + 12x + 9

= (2x + 3)²

i) 11x + 11y - x² - xy

= 11(x + y) - x(x + y)

= (!1 - x)(x + y)

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

b, \(x^2-2x-15=\left(x^2-2x+1\right)-16=\left(x-1\right)^2-4^2=\left(x-5\right)\left(x+3\right)\)

c, \(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)

d, \(12x^2y-18xy^2-30y^2=3y\left(4x^2-6xy-10y\right)\)

\(=3y\left[2y\left(2y-3x-5\right)\right]=6y^2\left(2y-3x-5\right)\)

e, \(5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\)

g, \(36-12x+x^2=\left(6-x\right)^2\)

h, \(4x^2+12x+9=\left(2x+3\right)^2\)

i, \(11x+11y-x^2-xy=11\left(x+y\right)-x\left(x+y\right)=\left(11-x\right)\left(x+y\right)\)

Mk sửa đề:x²-4y²+x+2y=(x-2y)(x+2y)+(x+2y)

                                      =(x+2y)(x=2y+1)

                                                          HOK TỐT NHA BN