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\(y-x^2y-2xy^2-y^3\)
= \(-y\left(x^2-2xy+y^2-1\right)\)
= \(-y\left[\left(x-y\right)^2-1\right]\)
= \(-y\left(x-y-1\right)\left(x-y+1\right)\)
y-x2y-2xy2-y3
=y(1-x2-2xy-y2)
=y[1-(x2+2xy+y2)]
=y[1-(x+y)2]
=y(1-x-y)(1+x+y)
\(a.=-4\left(x^2-2x+1\right)\)
\(=-4\left(x-1\right)^2\)
\(b.=3x\left(x-2\right)+2y\left(x-2\right)\)
\(=\left(3x+2y\right)\left(x-2\right)\)
2xy - x2 - y2 + 36=-(x2-2xy+y2-36)=-(x-y)2-36=-(x-y-6)(x-y+6)
a) 36 - 4a2 + 20ab - 25b2 = 36 - ( 4a2 - 20ab + 25b2 ) = 62 - ( 2a - 5b )2 = ( 6 - 2a + 5b )( 6 + 2a - 5b )
b) ( xy + 4 )2 - 4( x + y )2 = ( xy + 4 )2 - 22( x + y )2 = ( xy + 4 )2 - [ 2( x + y ) ]2
= ( xy + 4 )2 - ( 2x + 2y )2 = ( xy + 4 - 2x - 2y )( xy + 4 + 2x + 2y )
= [ x( y - 2 ) - 2( y - 2 ) ][ x( y + 2 ) + 2( y + 2 ) ]
= ( y - 2 )( x - 2 )( y + 2 )( x + 2 )
c) x2 + y2 - x2y2 + xy - x - y
= ( x2 - x2y2 ) + ( y2 - y ) + ( xy - x )
= x2( 1 - y2 ) + y( y - 1 ) + x( y - 1 )
= x2( 1 - y )( 1 + y ) - y( 1 - y ) - x( 1 - y )
= ( 1 - y )[ x2( 1 + y ) - y - x ) ]
= ( 1 - y )( x2 + x2y - y - x )
= ( 1 - y )[ ( x2 - x ) + ( x2y - y ) ]
= ( 1 - y )[ x( x - 1 ) + y( x2 - 1 ) ]
= ( 1 - y )[ x( x - 1 ) + y( x - 1 )( x + 1 ) ]
= ( 1 - y )( x - 1 )[ x + y( x + 1 ) ]
= ( 1 - y )( x - 1 )( x + xy + y )
d) 3x + 3y - x2 - 2xy - y2
= 3( x + y ) - ( x2 + 2xy + y2 )
= 3( x + y ) - ( x + y )2
= ( x + y )( 3 - x - y )
e) ( 2xy + 1 )2 - ( 2x + y )2
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )( y + 1 )( 2x + 1 )
a) \(36-4a^2+20ab-25b^2\)
\(=36-\left(4a^2-20ab+25b^2\right)\)
\(=36-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b) \(\left(xy+4\right)^2-4\left(x+y\right)^2\)
\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)
\(=\left(x+2\right)\left(x-2\right)\left(y+2\right)\left(y-2\right)\)
c) \(x^2+y^2-x^2y^2+xy-x-y\)
\(=-\left(x^2y^2-x^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=-x^2\left(y-1\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=\left(y-1\right)\left(-x^2y-x^2+y+x\right)\)
\(=\left(1-y\right)\left[\left(x^2y-y\right)+\left(x^2-x\right)\right]\)
\(=\left(1-y\right)\left(x-1\right)\left(xy+y+x\right)\)
tuổi con HN là :
50 : ( 1 + 4 ) = 10 ( tuổi )
tuổi bố HN là :
50 - 10 = 40 ( tuổi )
hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi
ta có sơ đồ : bố : |----|----|----|
con : |----| hiệu 30 tuổi
tuổi con khi đó là :
30 : ( 3 - 1 ) = 15 ( tuổi )
số năm mà bố gấp 3 tuổi con là :
15 - 10 = 5 ( năm )
ĐS : 5 năm
mình nha
\(x^2-y^2-2xy+y^2\)
\(=x^2-2xy+y^2-y^2\)
\(=\left(x-y\right)^2-y^2\)
\(=\left(x-y-y\right)\left(x-y+y\right)\)
\(=\left(x-2y\right)\left(x\right)\)
Giải
\(\left(x^2-2xy+y^2\right)-z^2\)
\(=\left(x-y\right)^2-z^2\)(Sử dụng hằng đẳng thức \(A^2-2AB+B^2=\left(A-B\right)^2\))
\(=\left(x-y+z\right)\left(x-y-z\right)\)(Sử dụng hằng đẳng thức \(A^2-B^2=\left(A+B\right)\left(A-B\right)\))
Vậy \(\left(x^2-2xy+y^2\right)-z^2=\left(x-y+z\right)\left(x-y-z\right)\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
Học tốt
\(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
a) x2 - y2 + 4x + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
b) x2 - 2xy + y2 - 1
= ( x2 - 2xy + y2 ) - 1
= ( x - y )2 - 12
= ( x - y - 1 )( x - y + 1 )
c) x2 - 2xy + y2 - 4
= ( x2 - 2xy + y2 ) - 4
= ( x - y )2 - 22
= ( x - y - 2 )( x - y + 2 )
d) x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z )( x - y + z )
e) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 52 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
f) x2 + y2 - 2xy - 4z2
= ( x2 - 2xy + y2 ) - 4z2
= ( x - y )2 - ( 2z )2
= ( x - y - 2z )( x - y + 2z )
\(x^2-2xy+y^2+3x-3y-10=\left(x^2-2xy+y^2\right)+3\left(x-y\right)-10=\left(x-y\right)^2+3\left(x-y\right)-10=\left(x-y\right)^2-2\left(x-y\right)+5\left(x-y\right)-10=\left(x-y\right)\left(x-y-2\right)+5\left(x-y-2\right)=\left(x-y+5\right)\left(x-y-2\right)\)
= ( x - y)^2 - 3 ( x - y) . -10
= ( x - y)^2 - 2.(x-y) . 3/2 +9/4 - 49/4
= ( x - y - 3/2) ^2 - (7/2)^2
= ( x- y - 3/2 - 7/2 )( x - y -3/2 + 7/2 )
=( x - y - 5 )( x - y + 2)
\(x^3-4x-12+3x^2=x\left(x^2-2^2\right)+3\left(x^2-2^2\right)=\left(x-2\right)\left(x+2\right)\left(x+3\right)\)
\(x^2+2xy-15y^2=x^2+2xy+y^2-16y^2=\left(x+y\right)^2-\left(4y\right)^2=\left(x-3y\right)\left(x+5y\right)\)
\(\left(x-y\right)^2-6\left(x-y\right)-16=\left(x-y\right)^2-2\times\left(x-y\right)\times3+9-25=\left(x-y-3\right)^2-5^2=\left(x-y-8\right)\left(x-y+2\right)\)
\(2xy-x^2-y^2+36\)
\(=-\left(x^2+2xy-y^2\right)+36\)
\(=-\left(x^2-2xy+y^2\right)+36\)
\(=-\left(x-y\right)^2+6^2\)
\(=-\left(x-y+6\right).\left(x-y-6\right)\)
Ta có: \(2xy-x^2-y^2+36=-\left(x^2-2xy+y^2-36\right)\)
\(=-\left[\left(x-y\right)^2-36\right]=-\left(x-y-6\right)\left(x-y+6\right)\)