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8 tháng 9 2020
A = xy + y - 2x - 2
= y( x + 1 ) - 2( x + 1 )
= ( x + 1 )( y - 2 )
B = x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
C = 3x2 - 3xy - 5x + 5y
= 3x( x - y ) - 5( x - y )
= ( x - y )( 3x - 5 )
D = xy + 1 + x + y
= y( x + 1 ) + ( x + 1 )
= ( x + 1 )( y + 1 )
E = ax - bx + ab - x2
= ( ax - x2 ) + ( ab - bx )
= x( a - x ) + b( a - x )
= ( a - x )( x + b )
F = x2 + ab + ax + bx
= ( ax + x2 ) + ( ab + bx )
= x( a + x ) + b( a + x )
= ( a + x )( x + b )
G = a3 - a2x - ay + xy
= a2( a - x ) - y( a - x )
= ( a - x )( a2 - y )
Bonus : = ( a - x )[ a2 - ( √y )2 ]
= ( a - x )( a - √y )( a + √y )
H = 2xy + 3z + 6y + xz
= ( 6y + 2xy ) + ( 3z + xz )
= 2y( 3 + x ) + z( 3 + x )
= ( 3 + x )( 2y + z )
XO
8 tháng 9 2020
A = xy + y - 2x - 2 = y(x + 1) - 2(x + 1) = (y - 2)(x + !1
B = x2 - 3x + xy - 3y = x(x - 3) + y(x - 3) = (x + y)(x - 3)
C = 3x2 - 3xy - 5x + 5y = 3x(x - y) - 5(x - y) = (3x - 5)(x - y)
D = xy + 1 + x + y = xy + x + y + 1 = x(y + 1) + (y + 1) = (x + 1)(y + 1)
E = ax - bx + ab - x2 = ax - x2 + ab - bx = a(a - x) - b(a - x) = (a - b)(a - x)
F = x2 + ab + ax + bx = ab + ax + bx + x2 = a(b + x) + x(b + x) = (a + x)(b + x)
G = a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
H = 2xy + 3z + 6y + xz = 2xy + 6y + 3z + xz = 2y(x + 3) + z(x + 3) = (2y + z)(x + 3)
12 tháng 10 2018
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
12 tháng 10 2018
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
25 tháng 2 2018
\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)..\)
\(=a^2\left(1-b\right)\left(1+b\right)-b\left(1-b\right)-a\left(1-b\right).\)
\(=\left(a^2+a^2b-b-a\right)\left(1-b\right)\)
\(=\left(ab+a+b\right)\left(a-1\right)\left(1-b\right)\)
9 tháng 7 2019
\(a^2+b^2-a^2b^2+ab-a-b\)
\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)\)
\(=a^2\left(1-b\right)\left(1+b\right)-b\left(1-b\right)-a\left(1-b\right)\)
\(=\left(a^2+a^2b-b-a\right)\left(1-b\right)\)
\(=\left(ab+a+b\right)\left(a-1\right)\left(1-b\right)\)
2T
31 tháng 8 2019
Câu a) dễ, ko làm
b) \(x^2y^2+1-x^2-y^2\)
\(=x^2\left(y^2-1\right)-\left(y^2-1\right)\)
\(=\left(x^2-1\right)\left(y^2-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
Câu c) đề sai
2T
31 tháng 8 2019
Câu c) ,đề đúng nek
\(bc\left(b+c\right)+ac\left(c-a\right)-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left[\left(b+c\right)-\left(a+b\right)\right]-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left(b+c\right)-ac\left(a+b\right)-ab\left(a+b\right)\)
\(=\left(b+c\right)\left(bc+ac\right)-\left(a+b\right)\left(ac+ab\right)\)
\(=\left(b+c\right)c\left(a+b\right)-\left(a+b\right)a\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\left(c-a\right)\)
18 tháng 8 2020
a)
\(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
b)
\(=a\left(a-b\right)+a-b\)
\(=\left(a+1\right)\left(a-b\right)\)
c)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left(x+1-y\right)\left(x+1+y\right)\)
d)
\(=x^3\left(x-2\right)+10x\left(x-2\right)\)
\(=x\left(x^2+10\right)\left(x-2\right)\)
e)
\(=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)
f)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
F
18 tháng 8 2020
a,2x3+3x2+2x+3
=(2x3+2x)+(3x2+3)
=2x(x2+1)+3(x2+1)
=(x2+1)(2x+3)
b,a2-ab+a-b
=(a2-ab)+(a-b)
=a(a-b)+(a-b)
=(a-b)(a+1)
c,2x2+4x+2-2y2
=2(x2+2x+1-y2)
=2[(x2+2x+1)-y2 ]
=2[(x+1)2-y2 ]
=2(x+1-y)(x+1+y)
d,x4-2x3+10x2-20x
=(x4-2x3)+(10x2-20x)
=x3(x-2)+10x(x-2)
=(x-2)(x3+10x)
=(x-2)[x(x2+10)]
e,x3+2x2+x
=x(x2+2x+1)
=x(x+1)2
f,xy+y2-x-y
=(xy+y2)-(x-y)
=y(x+y)-(x+y)
=(x+y)(y-1)
NH
1
15 tháng 7 2016
\(ab\left(x^2-y^2\right)-xy\left(a^2-b^2\right)\)
\(=abx^2-aby^2-xya^2+b^2xy\)
\(=\left(abx^2-xya^2\right)+\left(b^2xy-aby^2\right)\)
\(=ax\left(bx-ay\right)+by\left(bx-ay\right)\)
\(=\left(ax+by\right)\left(bx-ay\right)\)
TL:
\(ab\left(a^2+b^2\right)-xy\left(a^2+b^2\right)\)
\(=\left(ab-xy\right)\left(a^2+b^2\right)\)
ab(a2+b2)-xy(a2+b2)
=(ab-xy)(a2+b2)