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a)\(x^2+2xy-6y-9=x^2-9+2y.\left(x-3\right)=\left(x-3\right)\left(x+3\right)+2y.\left(x-3\right)\)
\(=\left(x-3\right).\left(x+3+2y\right)\)
b) \(x^3+x^2-6x=x^3-6x+9+x^2-9\)
\(=x^3+3x^2-3x^2-9x+3x+9+\left(x-3\right)\left(x+3\right)\)
\(=x^2.\left(x+3\right)-3x.\left(x+3\right)+3.\left(x+3\right)+\left(x-3\right).\left(x+3\right)\)
\(=\left(x+3\right).\left(x^2-3x+3+x-3\right)\)
\(=\left(x+3\right).\left(x^2-2x\right)=x.\left(x+3\right).\left(x-2\right)\)
a, \(x^2+2xy-6y-9\)
\(=\left(x^2+2xy+y^2\right)-\left(y^2+6y+9\right)\)
\(=\left(x+y\right)^2-\left(y-3\right)^2\)
\(=\left(x+2y-3\right)\left(x+3\right)\)
b, \(x^3+x^2-6x\)\(=x\left(x^2+3x-2x-6\right)\)
\(=x\left(x+3\right)\left(x-2\right)\)
câu a là x2-32 +2xy-6y= [x-3].[x+3]+2y.[x-3]=[x-3].[x+3+2y] câu b là x.[x2 +x-6]=x.[x2-22+x-2]=x.[x-2].[x+3]
mik nghĩ là sai đề r
X2 + 4y2 + 2( 3x + 6y + 2xy) +9
=x^2 +4y^2 +6x+ 12y+4xy+9
=x^2 +4xy+(2y)^2 +6(x+2y)+9
=(x+2y)^2+6(x+2y)+9
=(x+2y)(x+2y+6)+9
như vậy thì số 9 sẽ bị lẻ
\(a.=x^2-2x\)
\(b.=x\left(x+2y\right)-3\left(x+2y\right)\)
\(=\left(x-3\right)\left(x+2y\right)\)
\(c.=x\left(x^2-x-12\right)\)
\(=x\left(x^2-4x+3x-12\right)\)
\(=x \left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=x\left(x+3\right)\left(x-4\right)\)
x^2-y^2+10x-6y-9
=x^2+10x -(y^2+6y+9)
= x^2+10x-(y^2+2.3.y+3^2)
=x^2+10x-(y+3)^2
=\(\left[x^2-\left(y+3\right)^2\right]+10x \)
={\(\left(x+x+3\right)\left[x-\left(x+3\right)\right]\)}\(+10x\)
=\(\left(x+x+3\right).\left(x-x-3\right)+10x\)
(đến đây b tự lm nhé, m cx k bt đúng k nx )
a, \(x^3-6x^2+9x\)
\(=x\left(x^2-6x+9\right)\)
\(=x \left(x-3\right)\)
Câu b, c cũng tượng tự nha bn , dễ mà
#hoc_tot#
b) \(x^2-2xy+3x-6y=x\left(x-2y\right)+3\left(x-2y\right)=\left(x-2y\right)\left(x+3\right)\)
c)\(x^2-8x+7=x^2-x-7x+7=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\)
a)\(x^3-6x^2+9x=x\left(x^2-2\cdot x\cdot3+3^2\right)=x\left(x-3\right)^2\)
~ Chúc bạn học tốt ~
a) \(2x-6y=2\left(x-3y\right)\)
b) \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
c) \(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
d) \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e) \(x^3-10x^2+25x=x\left(x^2-10x+25\right)=x\left(x-5\right)^2\)
f) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
a. 2x - 6y
Bài làm
a. 2x - 6y
= 2(x - 3y)
b. x2 - y2
= (x - y)(x + y)
c. 2x3 + 4x2 + 2x
= 2x(x2 + 2 + 1)
= 2x(x2 + 3)
d. x2 - 2xy + y2 - 9
= (x2 - 2xy + y2) - 9
= (x - y)2 - 9
= (x - y - 3)(x - y + 3)
e. x3 - 10x2 + 25x
= x(x2 - 10x + 25)
= x(x2 - 2.5.x + 52)
= x(x + 5)2
f. xy + y2 - x - y
= y(x + y) - (x + y)
= (x + y)(y - 1)
a) 2x - 6y = 2( x - 3y )
b) x2 - y2 = ( x - y )( x + y )
c) 2x3 + 4x2 + 2x = 2x( x2 + 2x + 1 ) = 2x( x + 1 )2
d) x2 - 2xy + y2 = ( x2 - 2xy + y2 ) - 9 = ( x - y )2 - 32 = ( x - y - 3 )( x - y + 3 )
e) x3 - 10x2 + 25x = x( x2 - 10x + 25 ) = x( x - 5 )2
f) xy + y2 - x - y = ( xy + y2 ) - ( x + y ) = y( x + y ) - ( x + y ) = ( x + y )( y - 1 )
\(x^2+4x-9y^2+4\)
\(=\left(x^2+2.2x+2^2\right)-\left(3y\right)^2\)
\(=\left(x+2\right)^2-\left(3y\right)^2\)
\(=\left(x+2-3y\right)\left(x+2+3y\right)\)
\(x^2-9y^2-6y-1\)
\(=x^2-\left[\left(3y\right)^2+2.3y+1^2\right]\)
\(=x^2-\left(3y+1\right)^2\)
\(=\left(x-3y-1\right)\left(x+3y+1\right)\)
Tham khảo nhé~
a/Ta có:x2+4x-9y2+4
=x2+4x+4-(3y)2
=(x+2)2-(3y)2
=(x+2-3y)(x+2+3y)
b/Ta có:x2-9y2-6xy-1
=x2-6xy-(3y)2-1
=(x-3y-1)(x-3y+1)