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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 )
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)
b) \(5x-5y+ax-ay \)
\(=\left(5x-5y\right)+\left(ax-ay\right)\)
\(=5.\left(x-y\right)+a.\left(x-y\right)\)
\(=\left(x-y\right)\left(5+a\right)\)
c) \(a^3-a^2x-ay+xy\)
\(=\left(a^3-a^2x\right)-\left(ay-xy\right)\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a-x\right)\left(a^2-y\right)\)
a) \(A=5\left(x-y\right)+ax-ay=\left(a+5\right)\left(x-y\right)\)
b) \(B=a\left(x+y\right)-4x-4y=\left(x+y\right)\left(a-4\right)\)
c) \(C=xz+yz-5\left(x+y\right)=\left(x+y\right)\left(z-5\right)\)
d) \(D=a\left(x-y\right)+bx-by=\left(a+b\right)\left(x-y\right)\)
e) \(E=x\left(x+y\right)-5x-5y=\left(x-5\right)\left(x+y\right)\)
f) \(F=x^2-x-y^2-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
g) \(G=x^2-xy+x-y=x\left(x-y\right)+x-y=\left(x+1\right)\left(x-y\right)\)
A = 5(x - y) + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)
B = a(x + y) - 4x - 4y = a(x + y) - 4(x + y) = (a - 4)(x + y)
C = xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (z - 5)(x + y)
D = a(x - y) + bx - by = a(x - y) + b(x - y) = (a + b)(x - y)
E = x(x + y) - 5x - 5y = x(x + y) - 5(x + y) = (x - 5)(x + y)
F = x2 - x - y2 - y = (x2 - y2) - (x + y) = (x2 - xy + xy - y2) - (x + y) = [x(x - y) + y(x - y)] - (x + y) = (x - y)(x + y) - (x + y) = (x + y)(x - y - 1)
G = x2 - xy + x - y = x(x - y) + (x - y) = (x + 1)(x - y)
a) \(x^3-2x^2+2x-1^3\)
\(=x\left(x^2-2x+1\right)+x-1\)
\(=x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\)
b) \(x^2y+xy+x+1\)
\(=xy\left(x+1\right)+\left(x+1\right)\)
\(=\left(xy+1\right)\left(x+1\right)\)
c) \(ax+by+ay+bx\)
\(=a\left(x+y\right)+b\left(x+y\right)\)
\(=\left(a+b\right)\left(x+y\right)\)
d) \(x^2-\left(a+b\right)x+ab\)
\(=x^2-ax-bx+ab\)
\(=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
e) Ko biết làm
f) \(ax^2+ay-bx^2-by\)
\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
a) \(x^2-x-y^2+y\)
\(=x\left(x-1\right)-y\left(y-1\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
b) \(x^2-2xy-z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
c) \(5x-5y+ax-ay\)
\(=5\left(x-y\right)+a\left(x-y\right)\)
\(=\left(x-y\right)\left(5+a\right)\)
a) \(2bx-3ay-6by+ax\)
\(=2b\left(x-3y\right)+a\left(-3y+x\right)\)
\(=\left(2b+a\right)\left(x-3y\right)\)
b) \(x-2ac\left(x-y\right)-y\)
\(=\left(x-y\right)-2ac\left(x-y\right)\)
\(=\left(x-y\right)\left(1-2ac\right)\)
c) \(xy^2-by^2-ax+ab+y^2-a\)
\(=y^2\left(x-b\right)-a\left(x-b\right)+\left(y^2-a\right)\)
\(=\left(y^2-a\right)\left(x-b\right)+\left(y^2-a\right)\)
\(=\left(y^2-a\right)\left(x-b+1\right)\)
Bài 2
\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2x+6-x^2-3x=0\)
\(\Leftrightarrow-x^2-x+6=0\)
\(\Leftrightarrow-x^2-x=-6\)
\(\Leftrightarrow-x\left(x+1\right)=-6\)
\(\Rightarrow x=-3;x=2\)
Vậy \(x=-3\)và \(x=2\)
Bạn ơi mình nha
\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\)
\(10x^2+10xy+5x+5y=10x\left(x+y\right)+5\left(x+y\right)=5\left(2x+1\right)\left(x+y\right)\) \(5ay-3bx+ax-15by=a\left(5y+x\right)-3b\left(5y+x\right)=\left(a-3b\right)\left(5y+x\right)\) \(x^3+x^2-x-1=x^2\left(x+1\right)-\left(x+1\right)=\left(x^2-1\right)\left(x+1\right)=\left(x+1\right)^2\left(x-1\right)\) \(2bx-3ay-6by+ax=x\left(2b+a\right)-3y\left(2b+a\right)=\left(x-3y\right)\left(2b+a\right)\)
\(x+2a\left(x-y\right)-y=\left(x-y\right)+2a\left(x-y\right)=\left(1+2a\right)\left(x-y\right)\)