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

17 tháng 8 2021

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

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

17 tháng 8 2021

h, \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(\Rightarrow\left(27x^3-8\right):\left(3x-2\right)\\ =\left(3x-2\right)\left(9x^2+6x+4\right):\left(3x-2\right)\\ =9x^2+6x+4\)

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

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

26 tháng 5 2018

cảm ơn bn nhiều!!!!

26 tháng 5 2018

4. 4x2 + 4x + 1 = ( 2x + 1)2

5. \(\dfrac{1}{4}x-\dfrac{2}{3}xy+\dfrac{4}{9}y^2\) \(=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.\dfrac{2}{3}+\left(\dfrac{2}{3}y\right)^2\)

\(=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)^2\)

6. \(4a^2-\dfrac{4}{3}ab+\dfrac{1}{9}b^2=\left(2a\right)^2-2.2a.\dfrac{1}{3}+\left(\dfrac{1}{3}b\right)^2=\left(2a-\dfrac{1}{3}b\right)^2\)

7.

\(9x^2+4xy+\dfrac{4}{9}y^2-25z^2=\left(3x+\dfrac{2}{3}y\right)^2-\left(5z\right)^2=\left(3x+\dfrac{2}{3}y-5z\right)\left(3x+\dfrac{2}{3}y+5z\right)\)

26 tháng 5 2018

x2 - 6x + 9 = (x-3)2

x2 - 2x + 1 - y2 = ( x-1)2 - y2 = ( x-1-y)(x-1+y)

4x2 +12xy +9y2 = ( 2x+3y)2

30 tháng 7 2021

m, \(x^2+4x+4-4y^2=\left(x+2\right)^2-\left(2y\right)^2=\left(x+2-2y\right)\left(x+2+2y\right)\)

n, \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-\left(2z\right)^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)

24 tháng 12 2020

Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

\(=15x^4-12x^3+9x^2\)

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

31 tháng 7 2021

\(1,x^2-y^2+4x-4y\)

\(\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)

\(\left(x-y\right)\left(x+y+4\right)\)

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

\(\left(x-2y\right)\left(x+2y\right)+2\left(x-2y\right)\)

\(\left(x-2y\right)\left(x+2y+2\right)\)

\(3,3x^2-4y+4x-3y^2\)

\(3\left(x^2-y^2\right)-4\left(x-y\right)\)

\(3\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)

\(\left(x-y\right)\left(3x+3y-4\right)\)

\(x^4-6x^3+54x-81\)

\(x^4+3x^3-9x^3+27x^2-27x^2+81x-27x-81\)

\(\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27x+81\right)\)

\(x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)

\(\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)

\(\left(x+3\right)\left(x-3\right)^3\)

4 tháng 9 2020

1/   \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2.3xy+\left(3y\right)^2\)

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

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

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

Làm tạm 2 phần đợi mik xíu

4 tháng 9 2020

4x2 - 12xy + 9y2 = ( 2x )2 - 2.2x.3y + ( 3y )2 = ( 2x - 3y )2

x3 - y6 = x3  - ( y)3 = ( x - y2 )( x2 + xy2 + y4 )

x6 - 6x4 + 12x2 - 8 = ( x2 )3 - 3.(x2)2.2 + 3.x2.22 - 23 = ( x2 - 2 )3

( x2 + 4y2 - 5 )2 - 16( x2y2 + 2xy + 1 ) = ( x2 + 4y2 - 5 )2 - 42( xy + 1 )2

                                                            = ( x2 + 4y2 - 5 )2 - ( 4xy + 4 )2

                                                            = [ ( x2 + 4y2 - 5 ) - ( 4xy + 4 ) ][ ( x2 + 4y2 - 5 ) + ( 4xy + 4 ) ]

                                                            = ( x2 + 4y2 - 5 - 4xy - 4 )( x2 + 4y2 - 5 + 4xy + 4 )

                                                            = [ ( x2 - 4xy + 4y2 ) - 9 ][ ( x2 + 4xy + 4y2 ) - 1 ]

                                                            = [ ( x - 2y )2 - 32 ][ ( x + 2y )2 - 12 ]

                                                            = ( x - 2y - 3 )( x - 2y + 3 )( x + 2y - 1 )( x + 2y + 1 )

( a + b )3 - ( a3 + b3 ) = a3 + 3a2b + 3ab2 + b3 - a3 - b3

                                  = 3a2b + 3ab2

                                  = 3ab( a + b )