Phân tích đa thức thành nhân tử:
a) 2x3- 3x2
b) x2+ 5xy + x + 5y
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\(a,=4\left(x-5y\right)\\ b,=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
a: =5x(x-y)-7(x-y)
=(x-y)(5x-7)
b: =x(x+2y)+(x+2y)
=(x+2y)(x+1)
c; =(x-3)^2-9y^2
=(x-3-3y)(x-3+3y)
a
\(5x^2-5xy+7y-7x\\ =5x\left(x-y\right)+7\left(y-x\right)\\ =5x\left(x-y\right)-7\left(x-y\right)\\ =\left(5x-7\right)\left(x-y\right)\)
b
\(x^2+2xy+x+2y\\ =x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+1\right)\left(x+2y\right)\)
c
\(x^2-6x-9y^2+9\\ =x^2-6x+9-\left(3y\right)^2\\ =x^2-2.x.3+3^2-\left(3y\right)^2\\ =\left(x-3\right)^2-\left(3y\right)^2\\ =\left(x-3-3y\right)\left(x-3+3y\right)\)
\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)
Xem lại đề ý b
a: \(16x^5-25x^3\)
\(=x^3\left(16x^2-25\right)\)
\(=x^3\left(4x-5\right)\left(4x+5\right)\)
b: \(x^2-2xy+y^2-16\)
\(=\left(x-y\right)^2-16\)
\(=\left(x-y-4\right)\left(x-y+4\right)\)
c: \(x^2-5y-xy+5x\)
\(=x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+5\right)\)
d: \(x^2\left(x^2+4\right)-x^2+4\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
\(a,=x\left(x+y\right)+5\left(x+y\right)=\left(x+5\right)\left(x+y\right)\\ b,=x\left(y-x\right)-3\left(y-x\right)=\left(x-3\right)\left(y-x\right)\\ c,=18x-4x^3=2x\left(9-2x^2\right)\\ d,=\left(x-2\right)^2-4y^2=\left(x-2y-2\right)\left(x+2y-2\right)\\ e,=x^2-x-9x+9=\left(x-1\right)\left(x-9\right)\\ f,=4x^2-6x+2x-3=\left(2x-3\right)\left(2x+1\right)\)
a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+1\right)\)
b: \(=\left(x-4\right)\left(x+3\right)\)
e: =(x+3)(x-2)
a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)
b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)
c) \(=\left(2x\right)^2-\left(x^2+1\right)^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=4xy\left(y-3x+2\right)\)
e) \(=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)
f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)
g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)
h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)
i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)
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)
\(a,=x^2(2x-3)\\ b,=x(x+5y)+(x+5y)=(x+5y)(x+1)\)