giúp mình vs mn ơi

a: \(12x^4y^3+12x^3y^3+3x^2y^3\)

\(=3x^2y^3\cdot4x^2+3x^2y^3\cdot4x+3x^2y^3\cdot1\)

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

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

b: \(x^4+xy^3-x^3y-y^4\)

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

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

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

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

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

Lời giải:
$3x^2+4y^2+12x+3y+5=0$

$\Leftrightarrow 3(x^2+4x+4)+4y^2+3y-7=0$

$\Leftrightarrow 3(x+2)^2+(2y+\frac{3}{4})^2-\frac{121}{16}=0$

$\Leftrightarrow 3(x+2)^2=\frac{121}{16}-(2y+\frac{3}{4})^2\leq \frac{121}{16}$

$\Rightarrow (x+2)^2\leq \frac{121}{48}< 4$

$\Rightarrow -2< x+2< 2$

$\Rightarrow -4< x< 0$

$\Rightarrow x\in \left\{-3; -2; -1\right\}$

Đê đây bạn thay giá trị $x$ vào pt ban đầu để tìm $y$ thôi.

b: \(x^2-6x+xy-6y\)

\(=x\left(x-6\right)+y\left(x-6\right)\)

\(=\left(x-6\right)\left(x+y\right)\)

c: \(2x^2+2xy-x-y\)

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

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

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

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

\(=3\left(x+y\right)\left(x-y-1\right)\)

=3(x²-y²-4x-4y)

=3.(x+y).(x-y-4)

Phép chia hết xảy ra khi \(\left\{{}\begin{matrix}n+1\le3\\n+3\le5\end{matrix}\right.\Leftrightarrow n\le2\)