x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3

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

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

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

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

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

Đa thức này ko phân tích thành nhân tử được

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

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

(x^2 +7x)-(y^2+7y)

=x(x+7)-y(y+7)

=(x+7)(y+7)(x-y)

\(x^6+x^4+x^2y^2+y^4-y^6\)

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

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

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

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

\(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)

dễ 3x5

15 kết bn với mk nhé và k luôn nha

x⁶​—y⁶+(x⁴+x² y²+y⁴)=(x²)³—(y²)³+(x⁴+x² y²+y⁴)

=(x​​² — y²)(x⁴+x² y²+y⁴)+(x⁴+x² y²+y⁴)

=(x​⁴+x² y²+y⁴)(x² — y²+1)

`#3107.101107`

`x(y - 1) + 3(y - 1)`

`= (x + 3)(y - 1)`

x(y-1)+3(y-1)

=(y-1)(x+3)

Giải thích: đặt y-1 ra làm chung .... đa thức còn x+3

\(x-\sqrt{x}-2\\ =x+\sqrt{x}-2\sqrt{x}-2\\ =\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)\\ =\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)

\(x^4-x^2+2x+2\)

\(=x^4-2x^3+2x^2+2x^3-4x^2+4x+x^2-2x+2\)

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

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

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

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

\(x^4-x^2+2x+2\)

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

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

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

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