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a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
\(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
hk tốt
^^
d) x^3 + 2x^2 + 3x + 1
Bó tay
e) x^2 - 2x - 4y^2 - 4y
= x^2 - 2x + 1 - 4y^2 - 4y - 1
= ( x + 1 )^2 - ( 4y^2 + 4y + 1 )
= ( x + 1 )^2 - ( 2y + 1 )^2
= ( x+ 1 - 2y - 1 )( x + 1 + 2y + 1 )
= ( x - 2y )(x + 2y +2 )
4x2 là gì
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-2\right)\left(x+2y\right)\)
hk tốt
^^
c) 2x^3y - 2xy^3 - 4xy^2 - 2xy
= 2xy ( x^2 - y^2 - 2y - 1 )
= 2xy ( x^2 - ( y^2 + 2y + 1 )
= 2xy ( x^2 - ( y + 1 )^2 )
= 2x ( x - y - 1 )( x + y + 1 )
sai bạn ơi !
đáp án là
= 2xy (x + y + 1) (x - y + 1)
that pun cho ban Nguyen Dieu Thao :((
a/ \(=3y^2-6y-2x+1\)
b/ \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
c/ \(=\left(2-x\right)^3\)
d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)
\(=xy\left(y+x+3\right)+x^2\left(y+x+3\right)-3x\left(y+x+3\right)\)
\(=\left(xy+x^2-3x\right)\left(y+x+3\right)=x\left(y+x-3\right)\left(y+x+3\right)\)
e/ \(=xy-x^2+2x-y^2+xy-2y\)
\(=x\left(y-x+2\right)-y\left(y-x+2\right)=\left(x-y\right)\left(y-x+2\right)\)
a) =(2x+3y-1)2
b)=-(x-1)3
c)=-(x3-6x2+12x-8)=-(x-2)3
d)x3 + 2x2y + xy2 – 9x
= x(x2 + 2xy + y2 -9)
= x[(x2 + 2xy + y2) - 32]
= x[(x + y)2 - 32]
= x (x + y – 3)(x + y + 3)
e) 2x-2y-x2+2xy-y2=2(x-y)-(x-y)2=(x-y)(2-x+y)
\(a)\) \(x^2-2x-4y^2-4y\)
\(=\)\(\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\)\(\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\)\(\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\)\(\left(x-2y-2\right)\left(x+2y\right)\)
\(=\)\(2\left(x-y\right)\left(x+2y\right)\)
Chúc bạn học tốt ~
a) Ta có x2 - 2x - 4y2 - 4y
= x2 - 2x + 1 - 4y2 - 4y - 1
= (x - 1)2 - (4y2 + 4y + 1)
= (x - 1)2 - (2y + 1)2
= (x - 1 - 2y - 1)(x - 1 + 2y + 1)
= (x - 2y - 1)(x + 2y)
a) 5x3 - 40 = 5( x3 - 8 ) = 5( x - 2 )( x2 + 2x + 4 )
b) x2z + 4xyz + 4y2z = z( x2 + 4xy + 4y2 ) = z( x + 2y )2
c) 4x2 - y2 - 6x + 3y = ( 4x2 - y2 ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )
d) x2 + 2x - 4y2 + 1 = ( x2 + 2x + 1 ) - 4y2 = ( x + 1 )2 - ( 2y )2 = ( x - 2y + 1 )( x + 2y + 1 )
e) 3x2 - 3y2 - 12x + 12y = 3( x2 - y2 - 4x + 4y ) = 3[ ( x2 - y2 ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )
f) x3 + 5x2 + 4x + 20 = x2( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x2 + 4 )
g) x3 - x2 - 25x + 25 = x2( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x2 - 25 ) = ( x - 1 )( x - 5 )( x + 5 )
a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)
c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)
\(=\left(x+1\right)^2-4y^2=\left(x+2y+1\right)\left(x-2y+1\right)\)
e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)
\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)
\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)
g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)
\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)
\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)
a) x^2 - 4 + ( x - 2 )^2
= ( x- 2 )(x + 2 ) + ( x- 2)^2
= ( x - 2 ) ( x + 2 + x - 2 )
= 2x (x-2)
b) x^3 - 2x^2 + x - xy^2
= x ( x^2 - 2x + 1 - y^2)
= x [ ( x - 1 )^2 - y^2 ]
= x(x - 1 - y)( x - 1 + y )
c) x^3 - 4x^2 - 12x + 27
= x^3 + 3x^2 - 7x^2 - 21x + 9x + 27
= x^2 ( x + 3 ) - 7x ( x+ 3 ) + 9(x + 3 )
Để hai lần nha
= ( x+ 3 )(x^2 - 7x + 9 )
\(x^2-4+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x-2\right)\)
hk tốt
^^