Question

[Vũ Ngọc Hải Vân]

bài 3 ; áp dụng hằng đẳng thức để thực hiện phép chia

a, ( 4x mũ 2 + 12xy + 9y mũ 2 ) : ( 2x + 3y )

d, ( x mũ 2 + 6xy + 9y mũ 2 ) : ( x + 3y )

e, ( 64y mũ 3 - 27 ) : ( 4y - 3 ) 

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)=\left(2x+3y\right)^2:\left(2x+3y\right)=2x+3y\)

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

e: \(\dfrac{64y^3-27}{4y-3}=\dfrac{\left(4y-3\right)\left(16y^2+12y+9\right)}{4y-3}=16y^2+12y+9\)

Answer 2

[ILoveMath]

a, \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)

\(\Rightarrow\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)\)

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

b,\(x^2+6xy+9y^2=\left(x+3y\right)^2\)

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

c, \(64y^3-27=\left(4y-3\right)\left(16y^2+12y+9\right)\)

\(\Rightarrow\left(64x^3-27\right):\left(4y-3\right)\\ =\left[\left(4y-3\right)\left(16x^2+12x+9\right)\right]:\left(4y-3\right)\\ =16x^2+12x+9\)