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1, \(y^2+\left(3b+2a\right)xy+6abx^2\)
\(=y^2+3bxy+2axy+6abx^2\)
\(=y\left(y+3bx\right)+2ax\left(y+3bx\right)\)
= \(\left(y+2ax\right)\left(y+3bx\right)\)
2, \(ab\left(x-y\right)^2+8ab\)
=\(ab\left(x^2-2xy+y^2\right)+8ab\)
=\(ab\left(x^2-2xy+y^2+8\right)\)
3, \(x^2-\left(2a+b\right)+2aby^2\)
=\(x^2-2axy-bxy+2aby^{2^{ }}\)
=\(\left(x-by\right)\left(x-2ay\right)\)
4, \(xy\left(a^2+2b^2\right)+ab\left(x^2+y^2\right)\)
=\(a^2xy+2x^2ab+y^2ab+2b^2xy\)
=\(\left(ã+yb\right)\left(ay+2xb\right)\)
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)
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
b: \(=ab^2+ac^2+abc+bc^2+ba^2+abc+a^2c+b^2c+abc\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)
a: \(=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=-x^2\left(y-1\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=\left(y-1\right)\left(-x^2y-x^2+y+x\right)\)
\(=\left(1-y\right)\left(x^2y+x^2-x-y\right)\)
\(=\left(1-y\right)\cdot\left[y\left(x-1\right)\left(x+1\right)+x\left(x-1\right)\right]\)
\(=\left(1-y\right)\left(x-1\right)\left(xy+y+x\right)\)
1) \(\left(a^2+4\right)^2-16a^2\)
\(=\left(a^2+4-4a\right)\left(a^2+4+4a\right)\)
\(=\left(a-2\right)^2\cdot\left(a+2\right)^2\)
2) \(\left(a^2+9\right)^2-36a^2\)
\(=\left(a^2+9-6a\right)\left(a^2+9+6a\right)\)
\(=\left(a-3\right)^2\cdot\left(a+3\right)^2\)
3) \(\left(a^2+4b^2\right)^2-16a^2b^2\)
\(=\left(a^2+4b^2-4ab\right)\left(a^2+4b^2+4ab\right)\)
\(=\left(a-2b\right)^2\cdot\left(a+2b\right)^2\)
4) \(36a^2-\left(a^2+9\right)^2\)
\(=\left(6a-a^2-9\right)\left(6a+a^2+9\right)\)
\(=\left(6a-a^2-9\right)\left(a+3\right)^2\)
5) \(100a^2-\left(a^2+25\right)^2\)
\(=\left(10a-a^2-25\right)\left(10a+a^2+25\right)\)
\(=\left(10a-a^2-25\right)\left(a+5\right)^2\)
Phân tích đa thức thành nhân tử ( phối hợp các phương pháp )
1) x2 - ( a + b )xy + aby2
\(=x^2-axy-bxy+aby^2\)
\(=(x^2-axy)-(bxy+aby^2)\)
\(=x(x-ay)-by(x+ay)\)
\(=(x-ay)(x-by)\)
2) x2 + ( 2a + b )xy + 2aby2
=x2 + 2axy + bxy + 2aby2
=(x2+ bxy) +(2axy+ 2aby2 )
=x(x+ by) +2ay(x+ by)
=(x+ by)(x+2ay)
\(ab\left(x^2-y^2\right)-xy\left(a^2-b^2\right)\)
\(=abx^2-aby^2-xya^2+b^2xy\)
\(=\left(abx^2-xya^2\right)+\left(b^2xy-aby^2\right)\)
\(=ax\left(bx-ay\right)+by\left(bx-ay\right)\)
\(=\left(ax+by\right)\left(bx-ay\right)\)
a, x2 + (a +b) xy + aby2
=\(x\left(x+ay\right)+by\left(x+ay\right)\)
=\(\left(x+ay\right)\left(x+by\right)\)