Question

[Hồng Chiên]

Phân tích đa thức thành nhân tử:

a,14x^2y-21xy^2=28x^2y^2

b,x(x+y)-5x-5y

c,10x(x-y)-8(y-x)

d,(3x+1)^2-(x+1)^2

e,x^3+y^3+z^3-3xyz

g,5x^2-10xy+5y^2-20z^2                                                    

h,x^3-x+3x^2y+3xy^2+y^3-y

i,x^2+7x-8

k,x^2+4x+3

l,16x-5x^2-3

m,x^4+4

n,x^3-2x^2+x-xy^2                                                                         

Answer 1

[Miamoto Shizuka]

a, \(14x^2y-21xy^2-28x^2y^2=7xy\left(2x-3y-4xy\right)\)

b,\(x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

c,\(10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x-8\right)\)

d,\(\left(3x+1\right)^2-\left(x+1\right)^2=9x^2+6x+1-x^2-2x-1=7x^2-4x=x\left(7x-4\right)\)

e,\(x^3+y^3+z^3-3xyz=\)$=x^3+3\mathrm{xy(}x+\mathrm{y)}+y^3+{\mathrm{z3}}^{}-3\mathrm{xyz}-3\mathrm{xy}(\mathrm{x\; +}\mathrm{y)}$
$=($x$+y{)}^3+z^3-3\mathrm{xy}(x+\mathrm{y\; +}\mathrm{z)}$
$=(x+y+z)(x^2+2\mathrm{xy}+y^2-\mathrm{xy}-\mathrm{xz}+z^2)-3\mathrm{xy}(\mathrm{x\; +}y+z)$
$=(x+y+z)(x^2+y^2+z^2-\mathrm{xy}-yx-\mathrm{zx})$

$g,$5x^2 - 10xy + 5y^2 - 20z^2

$\mathrm{5[(x^2\; -\; 2xy\; +\; y^2)\; -\; 4z^2]}$