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19 tháng 6 2015

a) x^2+2xy+y^2-16

=(x+y)2-16

=(x+y-4)(x+y+4)

b) 3x^2+5x-3xy-5y

=(3x2-3xy)+(5x-5y)

=3x(x-y)+5(x-y)

=(x-y)(3x+5)

c) 4x^2-6x^3y-2x^2+8x

ko bik hoặc sai đề

d) x^2-4-2xy+y^2

=(x-y)2-4

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

e) x^3-4x^2-12x+27

=sai đề

g) 3x^2-18x+27

=3(x2-6x+9)

=3(x-3)2

h) x^2-y^2-z^2-2yz

=x2-(y2+z2+2yx)

=x2-(y+z)2

=(x-y-z)(x+y+z)

k) 4x^2(x-6)+9y^2(6-x)

=4x2(x-6)-9y2(x-6)

=(x-6)(4x2-9y2)

=(x-6)(2x-3y)(2x+3y)

l)6xy+5x-5y-3x^2-3y^2

=(5x-5y)+(-3x2+6xy-3y2)

=5(x-y)-3(x2-2xy+y2)

=5(x-y)-3(x-y)2

=(x-y)(5-3(x-y))

=(x-y)(5-3x+3y)

19 tháng 6 2015

b)x2+2xy+y2-16=(x+y)2-42=(x+y+4)(x+y-4)

c)3x2+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)

d)4x2-6x3y-2x2+8x=2x(2x-3x2y-x+4)

e)x2-4-2xy+y2=(x2-2xy+y2)-4=(x-y)2-22=(x-y-2)(x-y+2)

k)x2-y2-z2-2yz=x2-(y+z)2=(x-y-z)(x+y+z)

m)6xy+5x-5y-3x2-3y2=3(x2-2xy+y2)+5(x-y)=3(x-y)2+5(x-y)=(x-y)(3x-3y+5)


 

27 tháng 6 2016

b. (x^2+2xy+y^2)-16 =(x+y)^2-16=(x+y+4)(x+y-4)

2 tháng 3 2020

1)2xy+3z+6y+xz 

= x(2y + z) + 3(z + 2y)

= (x + 3)(2y + z)

2)x^4-9x^3+x^2-9x 

= x^2(x^2 + 1) - 9x(x^2 + 1)

= (x^2 + 1)(x^2 - 9x)

= x(x - 9)(x^2 + 1)

3)x^2-xy+x-y 

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

= (x + 1)(x - y)

4)xz+yz-5(x+y)

= z(x + y) - 5(x + y)

= (z - 5)(x + y)

5)3x^2-3xy-5x+5y 

= 3x(x - y) - 5(x - y)

= (3x - 5)(x - y)

6)x^2+4x-y^2+4y 

= (x - y)(x + y) + 4(x + y)

= (x - y + 4)(x + y)

20 tháng 8 2021

a) \(3xy-6xy^2=3xy\left(1-2y\right)\)

b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

c) \(x^3-x^2+2\)

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

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

f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)

g) \(6x^2-12x=6x\left(x-2\right)\)

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

i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

20 tháng 8 2021

k) \(2x^3+2x^2y-4xy^2=2x\left(x^2+xy-2y^2\right)\)

l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)

6 tháng 9 2021

a) x2+y2-4x+4y+8=0

⇔ (x-2)2+(y+2)2=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x2-4xy+y2=0

⇔ x2+(2x-y)2=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x2+2y2+z2-2xy-2y-4z+5=0

⇔ (x-y)2+(y-1)2+(z-2)2=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

2 tháng 3 2020

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

2) 4x2 - 9y2 + 4x - 6y = (4x2 - 9y2) + (4x - 6y) = (2x - 3y)(2x + 3y) + 2(2x - 3y) = (2x - 3y)(2x + 3y + 2)

3) x2 + x + y2 + y + 2xy = (x2 + 2xy + y2) + (x + y) = (x + y)2 + (x + y) = (x + y)(x + y + 1)

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

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

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

7) x2 + 2xz - y2 + 2uy + z2 - u2 =(x2 + 2xz + z2) - (y2 - 2uy + u2) = (x + z)2 - (y - u)2 = (x + z - y + u)(x + z + y - u)

8) x3 + 3x2y + x + 3xy2 + y + y3 = (x3 + 3x2y + 3xy2 + y3) + (x + y) = (x + y)3 + (x + y) = (x + y)(x2 + 2xy + y2 + 1)

9) x3 + y(1 - 3x2) + x(3y2 - 1) - y3 = x3 + y - 3x2y + 3xy2 - x - y3 = (x3 - 3x2y + 3xy2 - y3) - (x - y) = (x - y)3 - (x - y) = (x - y)(x2 - 2xy+y2-1)

6 tháng 4 2020

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)

6 tháng 4 2020

Thank you

8 tháng 8 2019

1/x^3 - 2x^2 - 9x + 18

= x\(^2\)( x - 2 ) - 9 ( x - 2 ) = ( x\(^2\) - 9 ) ( x - 2 )= ( x - 3 ) ( x +3 ) ( x - 2 )

2/3x^2 -5x - 3y^2 + 5y

= 3( x\(^2\) - y\(^2\) ) - 5 ( x - y ) = 3 ( x - y ) ( x + y ) - 5 ( x - y ) = ( x - y ) [ 3( x+ y ) - 5 ]

= ( x - y ) ( 3x + 3y - 5 )

3/49 - x^2 + 2xy - y^2

= 49 - ( x\(^2\) - 2xy + y\(^2\) ) = 49 - ( x - y )\(^2\) = ( 7 - x + y ) ( 7 + x - y )

5/ x^2 - 4x^2y^2 + 2xy

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

6/ 3x - 3y - x^2 + 2xy - y^2

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