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a) 10x(x-y) - 6y(y-x)
= 10x(x-y) +6y ( x-y)
=(10x+6y) (x-y)
b) 3x2 + 5y - 3xy -5x
= 3x(x-y) + 5(y-x)
= 3x(x-y) -5(x-y)
= (3x-5) ( x-y)
c) 3y2 - 3z2 +3x2 + 6xy
=3(y2 - z2 + x2 + 2xy)
=3[(x2 +2xy+y2)-z2 ]
=3[(x+y)2 - z2 ]
=3(x+y-z) (x+y+z)
d) 16x3 + 54y3
=2(8x3 + 27y3 )
=2[(2x)3 + (3y)3 ]
=2(2x+3y) (4x2 - 6xy + 9y2 )
e) x2 - 25 -2xy+y2
=(x2-2xy+y2)-25
=(x-y)2 -52
=(x-y-5) (x-y+5)
f) (mình chưa làm ra )
{mong m.n bổ sung thêm..}
mấy câu trên bạn kia đã trả lời rồi nên mk k làm lại nx
f, x5 - 3x4 + 3x3 - x2
= x2 (x3 - 3x2 + 3x -1)
= x2 (x - 1)3
Chúc bạn học tốt!
a) \(=a\left(a^3-9a^2+a-9\right)=a\left[a^2\left(a-9\right)+\left(a-9\right)\right]\)
\(=a\left(a-9\right)\left(a^2+1\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(2y+z\right)+3\left(2y+z\right)=\left(2y+z\right)\left(x+3\right)\)
d) \(=x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-a\right)\left(x-b\right)\)
a) = a(a³-9a²+a-9)
b) =3x²+5y-3xy-5x
= (3x²-5x)+(5y-3xy)
=x(3x-5)+y(5-3x)
=x(3x-5)-y(3x-5)
=(3x-5)(x-y)
c)2xy +3z+6y+xz
=(2xy+6y)+(3z+xz)
=2y(x+3)+z(3+x)
=(x+3)(2y-z)
a) \(5x+10y=5\left(x+2y\right)\)
b) \(3x^3-12x=3x\left(x^2-4\right)=3x\left(x-2\right)\left(x+2\right)\)
c) \(4x^2+9x-4xy-9y=4x\left(x-y\right)+9\left(x-y\right)=\left(x-y\right)\left(4x+9\right)\)
d) \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)
\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)
\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)
\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)
a: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
3x2 + 6xy + 3y2 – 3z2
= 3.(x2 + 2xy + y2 – z2)
(Nhận thấy xuất hiện x2 + 2xy + y2 là hằng đẳng thức nên ta nhóm với nhau)
= 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2]
= 3(x + y – z)(x + y + z)
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)