Câu 1:

$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$

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

Câu 2:

$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$

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

Câu 1:

\(x^2+4y^2+4xy-16\)

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

\(=\left(x+2y+4\right)\left(x+2y-4\right)\)

Câu 2:

\(x^3+x^2+y^3+xy\)

\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)

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

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

C1:x^2+4y^2+4xy-16

=[x^2+4xy+(2y)^2]-16

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

=(x+2y-4)(x+2y+4)

C2: x^3+x^2+y^3+xy

=(x^2+xy)+(x^3+y^3)

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

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

bài này ra lâu r nhưng ngứa tay nên giải luôn=)))))

\(=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)

Ta có: \(x^3-\left(y-2\right)^3+\left(y-x-2\right)^2\)

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

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

\(=\left(x-y+2\right)\left(x^2+y^2+6+xy-x-5y\right)\)

\(x^3-y^3+2x^2+2xy\)

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

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

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

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

= x [ ( x^2 - y^2 ) + ( 2x + 2y ) ]

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

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

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

= x(x^2 + 2xy + y^2 - 25z^2)

= x(x + y - 5z)(x + y + 5z)

A = x.[x^2.(x^2-7)^2-36]

   = x.[(x^3-7x)^2-6^2]

   = x.(x^3-7x-6).(x^3-7x+6)

   = x.[(x^3+1)-(7x+7)].[(x^3-x)-(6x-6)]

   = x.(x+1).(x^2-x-7).(x-1).(x^2+x-6)

   = x.(x+1).(x-1).(x-2).(x+3).(x^2-x-7)

Tk mk nha

x3(x2−7)2−36x=x3(x4−14x2+49)−36xx3(x2−7)2−36x=x3(x4−14x2+49)−36x

=x7−14x5+49x3−36xx7−14x5+49x3−36x

=x7−x6+x6−x5−13x5+13x4−13x4+13x3+36x3−36xx7−x6+x6−x5−13x5+13x4−13x4+13x3+36x3−36x

=x6(x−1)+x5(x−1)−13x4(x−1)−13x3(x−1)+36x(x2−1)x6(x−1)+x5(x−1)−13x4(x−1)−13x3(x−1)+36x(x2−1)

=x(x−1)(x5+x4−13x3−13x2+36x+36)x(x−1)(x5+x4−13x3−13x2+36x+36)

=x(x−1)[x4(x+1)−13x2(x+1)+36(x+1)]x(x−1)[x4(x+1)−13x2(x+1)+36(x+1)]

=x(x−1)(x+1)(x4−13x2+36)x(x−1)(x+1)(x4−13x2+36)

đặt x^2 =a (a>=0) thì xét đa thức x4−13x2+36=a2−13a+36x4−13x2+36=a2−13a+36

xét Δ=b2−4ac=169−4.36=25Δ=b2−4ac=169−4.36=25

Δ>0Δ>0→phương trình có 2 nghiệm riêng biệt là ⎡⎣a1=−b+Δ√2a=13+52=9a2=−b−Δ√2a=13−52=4[a1=−b+Δ2a=13+52=9a2=−b−Δ2a=13−52=4(t/m a>=0)

vậy bt ban đầu :x(x−1)(x+1)(x2−4)(x2−9)x(x−1)(x+1)(x2−4)(x2−9)

=(x−3)(x−2)(x−1)x(x+1)(x+2)(x+3)

A = x.[x^2.(x^2-7)^2-36]

   = x.[(x^3-7x)^2-6^2]

   = x.(x^3-7x-6).(x^3-7x+6)

   = x.[(x^3+1)-(7x+7)].[(x^3-x)-(6x-6)]

   = x.(x+1).(x^2-x-7).(x-1).(x^2+x-6)

   = x.(x+1).(x-1).(x-2).(x+3).(x^2-x-7)