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A= \(^{x^3+3x^2y-4xy^2-12y^3=x^2\left(x+3y\right)-4y^2\left(x+3y\right)=\left(x+3y\right)\left(x^2-4y^2\right)}\)
a) \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
b) \(x^2-2xy-4+y^2\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
g) \(12xy-12xz+3x^2y-3x^2z\)
\(=12x\left(y-z\right)+3x^2\left(y-z\right)\)
\(=3x\left(4+x\right)\left(y-z\right)\)
e) \(16x^2-9\left(x^2+2xy+y^2\right)\)
\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)
\(=\left(4x-3\left(x+y\right)\right)\left(4x+3\left(x+y\right)\right)\)
\(=\left(x+y\right)\left(7x+y\right)\)
d) làm tương tự như phần g chỉ khác là phải nhóm( nhóm xen kẽ), phần f cũng vậy
a, 25-x2+4xy-4y2
= 25-(x2-4xy+4y2)
= 52-(x-2y)2
= (5-x+2y)(5+x-2y)
Các biểu thức sau bạn tự chứng minh nhé
a) 12xy( x2 - 2xy + y2) = 12xy( x - y )2
b) ( x2 + xy ) - ( 6x + 6y ) = x( x + y ) - 6( x + y )
= ( x + y )(x - 6)
c) ( 2x2 + 2xy ) - ( x + y ) = 2x(x + y ) - ( x + y )
= (x + y )(2x - 1)
e) ( 3x2 - 3y2 ) - ( 12x + 12y ) = 3( x2 - y2 ) - 12( x + y)
= 3(x - y)(x + y) - 12(x + y) = ( x + y )(3x - 3y - 12)
= 3( x + y )(x - y -4)
g) \(\left[x\left(x+10\right)\right].\left[\left(x+4\right)\left(x+6\right)\right]\) + 128
= (x2 + 10x).(x2 + 10x + 24) + 128
Đặt x2 + 10x + 12 = t
⇒ Biểu thức trên có dạng:
( t - 12 )(t + 12) + 128 = t2 - 144 + 128 = t2 - 16 = t2 - 42
= ( t - 4 )( t + 4) = (x2 + 10x + 12 - 4 )(x2 + 10x + 12 + 4)
= ( x2 + 10x + 8)(x2 + 10x + 16)
f) -2xy + 4y2 = 2y( -x + 2y)
Có 2 phần g nha bạn. Mk chuyển phần cuối thành phần f.
Phần d do mk hơi ngu nên chưa nghĩ ra bạn thông cảm nha.
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3xy\left(x-y\right)-12\left(x-y\right)\)
\(=\left(3xy-12\right)\left(x-y\right)\)
2) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
Ta có : 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 4.3.(x - y)
= 3(x - y)(x + y - 4)
c) \(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
b) \(x^3+3x^2-3x-1\)
\(=\left(x^3-1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
a) \(8a^2xy-18b^2xy=2xy\left(4a^2-9b^2\right)=2xy\left(2a-3b\right)\left(2a+3b\right)\)
b) \(32a^2b^2-4=4\left(8a^2b^2-1\right)\)
c) \(x^2-49z^2-4xy+4y^2=\left(x^2-4xy+4y^2\right)-49z^2\)
\(=\left(x-2y\right)^2-\left(7z\right)^2=\left(x-2y+7z\right)\left(x-2y-7z\right)\)
d) \(3x^2+6x+3-3y^2=3\left(x^2+2x+1-y^2\right)=3.\left[\left(x+1\right)^2-y^2\right]\)
\(=3\left(x-y+1\right)\left(x+y+1\right)\)
e) \(12x^2y-12y^3+36xy+27y=3y\left(4x^2-4y^2+12x+9\right)\)
\(=3y\left[\left(4x^2+12x+9\right)-4y^2\right]=3y\left[\left(2x+3\right)^2-\left(2y\right)^2\right]\)
\(=3y\left(2x-2y+3\right)\left(2x+2y+3\right)\)
a) 8a2xy - 18b2xy
= 2xy( 4a2 - 9b2 )
= 2xy( [ ( 2a )2 - ( 3b )2 ]
= 2xy( 2a - 3b )( 2a + 3b )
b) 32a2b2 - 4
= 4( 8a2b2 - 1 )
c) x2 - 49z2 - 4xy + 4y2
= ( x2 - 4xy + 4y2 ) - 49z2
= ( x - 2y )2 - ( 7z )2
= ( x - 2y - 7z )( x - 2y + 7z )
d) 3x2 + 6x + 3 - 3y2
= 3( x2 + 2x + 1 - y2 )
= 3[ ( x2 + 2x + 1 ) - y2 ]
= 3[ ( x + 1 )2 - y2 ]
= 3( x - y + 1 )( x + y + 1 )
e) 12x2y - 12y3 + 36xy + 27y
= 3y( 4x2 - 4y2 + 12x + 9 )
= 3y[ ( 4x2 + 12x + 9 ) - 4y2 ]
= 3y[ ( 2x + 3 )2 - ( 2y )2 ]
= 3y( 2x - 2y + 3 )( 2x + 2y + 3 )