Question

1) phân tích đa thức thành nhân tử

\(x^2-3x-4\)

2)Rút gọn rồi tính giá trị của biểu thức

\(\left(9x^2y^2-6x^2y^3+15xy\right)\):(-3xy) tại \(x=1,y=2\)

Answer 1

1/

x² - 3x - 4 

= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)

\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)

\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)

\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)

\(=\left(x-4\right)\left(x+1\right)\)

Answer 2

Bài 1 :

\(x^2-3x-4\)

\(=x^2+x-4x-4\)

\(=x\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x+1\right)\left(x-4\right)\)

Answer 3

a)

x^2 -3x -4

=x^2 -2.x.1,5 + 1,5^2 -6,25

=(x+1,5)^2 -2,5^2

=(x -1,5+ 2,5).(x-1,5-2,5)

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

Answer 4

2/

A = \(\left(9x^2y^2-6x^2y^3+15xy\right):\left(-3xy\right)\)

\(=3xy\left(3xy-2xy^2+5\right):\left(-3xy\right)\)

\(=\frac{3xy\left(3xy-2xy^2+5\right)}{-3xy}\)

\(=-\left(3xy-2xy^2+5\right)\)

\(=-3xy+2xy^2-5\)

\(=xy\left(-3+2y\right)-5\)

tại x = 1 , y = 2

\(A=1.2\left(-3+2.2\right)-5\)

\(=2.\left(-3+4\right)-5\)

\(=2-5\)

\(=-3\)

Answer 5

1) \(x^2-3x-4=x^2-x+4x-4=\left(x^2-x\right)+\left(4x-4\right)=x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x+4\right)\)

2)\(\left(9x^2y^2-6x^2y^3+15xy\right):\left(-3xy\right)=\frac{3xy\left(3xy-2xy^2+5\right)}{-3xy}=\frac{-3xy\left(3xy-2xy^2+5\right)}{3xy}=-3xy+2xy^2-5\)

Thay x=1,y=2 vào biểu thức trên ta được

\(-3.1.2+2.1.2^2-5=-6+8-5=-3\)

Answer 6

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

\(x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)

2