a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)

b, \(x^3+x^2-9x-9=0\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-9\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=-1;\pm3\)

c, \(x^2-3x-10=0\Leftrightarrow x^2+2x-5x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=5;-2\)

a, \(20x^2y^3-15xy^2=5xy^2\left(4xy-3\right)\)

b, \(3x+3y-x^2-xy=3\left(x+y\right)-x\left(x+y\right)=\left(3-x\right)\left(x+y\right)\)

c, \(9-x^2-y^2+2xy=9-\left(x^2+y^2-2xy\right)\)

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

20x²y³ - 15xy² = 5xy²( 4xy - 3 )

3x + 3y - x² - xy = ( 3x + 3y ) - ( x² + xy ) = 3( x + y ) - x( x + y ) = ( x + y )( 3 - x )

9 - x² - y² + 2xy = 9 - ( x² - 2xy + y² ) = 3² - ( x - y )² = ( 3 - x + y )( 3 + x - y )

là gì?

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)0a

Đặt f(x) = 2x² + ax + 1

      g(x) = x - 3

f(x) chia g(x) dư 4 

=> f(x) - 4 chia hết cho g(x)

<=> 2x² + ax + 1 - 4 chia hết cho x - 3

<=> 2x² + ax - 3 chia hết cho x - 3

Áp dụng định lí Bézout ta có :

f(x) - 4 chia hết cho g(x) <=> f(3) - 4 = 0

<=> 18 + 3a - 3 = 0

<=> 3a + 15 = 0

<=> 3a = -15

<=> a = -5

Vậy a = -5

Sửa đề : \(x^2-xz-9y^2+3yz=\left(x^2-9y^2\right)+\left(-xz+3yz\right)\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left[\left(x+3y\right)-z\right]=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=x^4+10x^3+24x^2+10x^3+100x^2+240x+128\)

\(=x^4+20x^3+124x^2+240x+128\)

hong biet=))