Question

Phân tích đa thức thành nhân tử:
a, A = x\(^3\)- 2x\(^2\)+ x
b, B = 2x\(^2\)+ 4x + 2 - 2y\(^2\)
c, C = 2xy - x\(^2\)- y\(^2\)+ 16
d, D = x\(^3\)+ 2x\(^2\)y + xy\(^2\)- 9x
e, E = 2x - 2y - x\(^2\)+ 2xy - y\(^2\)

Answer 1

A=\(x^3-2x^2+x\)

=x.(x²-2x+1)

=x(x-1)²

B=\(2x^2+4x+2-2y^2\)

=\(2\left(x^2+2x+1-y^2\right)\)

=\(2.\left[\left(x+1\right)^1-y^2\right]\)

=\(2\left(x+1-y\right)\left(x+1+y\right)\)

C=\(2xy-x^2-y^2+16\)

=\(-\left(-2xy+x^2+y^2-16\right)\)

=\(-\left[\left(x-y\right)^2-4^2\right]\)

=-(x-y-4)(x-y+4)

D=\(x^3+2x^2y+xy^2-9x\)

=\(x\left(x^2+2xy-y^2-9\right)\)

=\(x.\left[\left(x-y\right)^2-3^2\right]\)

=x.(x-y-3)(x-y+3)

E=\(2x-2y-x^2+2xy-y^2\)

\(=\left(2x-2y\right)-\left(x^2-2xy+y^2\right)\)

=\(2\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)

=(x-y)(2x-2y-x+y)

=(x-y)(x+y)

Answer 2

a, \(A=x^3-2x^2+x\)

\(=x^3-x^2-x^2+x\)

\(=x^2\left(x-1\right)-x\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-x\right)\)

\(=x\left(x-1\right)^2\)

b, \(B=2x^2+4x+2-2y^2\)

\(=2\left(x^2+2x+1\right)-2y^2\)

\(=2\left(x+1\right)^2-2y^2\)

\(=2\left(x+1+y\right)\left(x+1-y\right)\)

c, \(C=2xy-x^2-y^2+16\)

\(=-\left[\left(x^2-2xy+y^2\right)-16\right]\)

\(=-\left[\left(x-y\right)^2-16\right]\)

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

d, \(D=x^3+2x^2y+xy^2-9x\)

\(=x\left(x^2+2xy+y^2-9\right)\)

\(=x\left[\left(x+y\right)^2-3\right]\)

\(=x\left(x+y+3\right)\left(x+y-3\right)\)

e, \(E=2x-2y-x^2+2xy-y^2\)

\(=2\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(2-x+y\right)\)

Answer 3

a ) \(A=x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)

b ) \(B=2x^2+4x+2-2y^2=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x+1-y\right)\left(x+1+y\right)\)

c ) \(C=2xy-x^2-y^2+16=16-\left(x^2-2xy+y^2\right)=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x+y\right)\)

d ) \(D=x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-9\right]=x\left(x+y-3\right)\left(x+y+3\right)\)

e ) \(E=2x-2y-x^2+2xy-y^2=-\left(x-y\right)^2+2\left(x-y\right)\)

Đặt \(x-y=a\) , ta được :

\(-a^2+2a\)

\(=-\left(a^2-2a+1\right)+1\)

\(=-\left(a-1\right)^2+1\)

\(=\left(1-a+1\right)\left(1+a-1\right)\)

\(=\left[1-\left(x-y\right)+1\right]\left[1+x-y-1\right]\)

\(=\left(2-x+y\right)\left(x-y\right)\)

Answer 4

a) A= x.(x² - 2x + 1) = x.(x - 1 )²

b) B= 2.( x² + 2x + 1 - y² ) = 2.\(\left[\left(x+1\right)^2-y^2\right]\)

B = 2.(x + 1 - y ).(x + 1 + y)

c) C = - ( x² - 2xy + y² - 16) = \(\left[-\left(x-y\right)^2-4^2\right]\)

C = -( x- y-4).(x-y+4)

d) D = x.(x² + 2xy + y² - 9) = \(x.\left[\left(x+y\right)^2-3^2\right]\)

D = x.( x + y - 3 ).(x+ y + 3)

e) E = ( 2x - 2y ) - (x² - 2xy +y²) = 2.(x - y) - ( x - y )²

E = ( x - y).2.( x - y)