Question

Phân tích đa thức thành nhân tử bằng nhiều phương pháp

a,2x^3+16

b,8xy^3+x(x-y)^3

c,(a+b)^3+c^3

d,x^2+(a+b)xy+aby^2

e,x^2-(2a+b)xy+2aby^2

g,y^2-(3a+2b)xy+6abx^2

f,3xy(a^2+b^2)-ab(x^2+9y^2)

h,(xy+ab^2)+(ay-bx)^2

j,ab(x^2+y^2)+xy(a^2+b^2)

k,(xy-ab)^2+(bx+ay)^2

Answer 1

a)

$2x^3+16=2(x^3+8)=2(x^3+2^3)=2(x+2)(x^2-2x+4)$

b)

$8xy^3+x(x-y)^3=x[8y^3+(x-y)^3]=x[(2y)^3+(x-y)^3]$

$=x(2y+x-y)[(2y)^2-2y(x-y)+(x-y)^2]$

$=x(x+y)(x^2-4xy+7y^2)$

c)

$(a+b)^3+c^3=(a+b+c)[(a+b)^2-(a+b)c+c^2]$

$=(a+b+c)(a^2+b^2+c^2+2ab-ac-bc)$

d)

\(x^2+(a+b)xy+aby^2=x^2+axy+bxy+aby^2\)

$=x(x+ay)+by(x+ay)=(x+by)(x+ay)$

Answer 2

e)

$x^2-(2a+b)xy+2aby^2=x^2-2axy-bxy+2aby^2$

$=x(x-2ay)-by(x-2ay)=(x-by)(x-2ay)$

g)

$y^2-(3a+2b)xy+6abx^2=(y^2-2bxy)-(3axy-6abx^2)$

$=y(y-2bx)-3ax(y-2bx)=(y-3ax)(y-2bx)$

f)

$3xy(a^2+b^2)-ab(x^2+9y^2)=3xya^2+3xyb^2-abx^2-9aby^2$

$=(3xya^2-abx^2)-(9aby^2-3xyb^2)$

$=ax(3ay-bx)-3by(3ay-bx)=(3ay-bx)(ax-3by)$

Answer 3

h) Sửa lại đề bài chút xíu:

$(xy+ab)^2+(ay-bx)^2=x^2y^2+a^2b^2+2abxy+a^2y^2-2aybx+b^2x^2$

$=x^2y^2+a^2b^2+a^2y^2+b^2x^2$

$=(x^2y^2+b^2x^2)+(a^2b^2+a^2y^2)$

$=x^2(y^2+b^2)+a^2(b^2+y^2)=(a^2+x^2)(b^2+y^2)$

j)

$ab(x^2+y^2)+xy(a^2+b^2)=abx^2+aby^2+xya^2+xyb^2$
$=(abx^2+xya^2)+(aby^2+xyb^2)$

$=ax(bx+ay)+by(ay+bx)=(ax+by)(ay+bx)$

k)

$(xy-ab)^2+(bx+ay)^2=x^2y^2-2xyab+a^2b^2+b^2x^2+2bxay+a^2y^2$

$=x^2y^2+a^2b^2+b^2x^2+a^2y^2$

$=(x^2y^2+b^2x^2)+(a^2b^2+a^2y^2)=x^2(y^2+b^2)+a^2(b^2+y^2)$

$=(a^2+x^2)(b^2+y^2)$