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

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

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

\(12y-9x^2+36-3x^2y\)

\(=\left(36-9x^2\right)-\left(3x^2y-12y\right)\)

\(=-9\left(x^2-4\right)-3y\left(x^2-4\right)\)

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

\(12y-9x^2+36-3x^2y\)

\(=\left(12y-3x^2y\right)-\left(9x^2-36\right)\)

\(=3y\left(4-x^2\right)-9\left(x^2-4\right)\)

\(=3y\left(4-x^2\right)+9\left(4-x^2\right)\)

\(=\left(3y+9\right)\left(4-x^2\right)\)

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

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

=(−3)(x−2)(x+2)(3+y)

\(=x^2+2x\cdot\frac{1}{2}+\frac{1}{4}-\left(\frac{\sqrt{23}}{2}i\right)^2\)

\(=\left(x+\frac{1}{2}\right)^2\)\(-\left(\frac{\sqrt{23}}{2}i\right)^2\)

\(\left(x+\frac{1}{2}-\frac{\sqrt{23}}{2}i\right)\left(x+\frac{1}{2}+\frac{\sqrt[]{23}}{2}i\right)\)

\(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)

\(\dfrac{1}{4}x^2+2xy+4y^2=\left(\dfrac{1}{2}x+2y\right)^2\)

Ok

Mình không giảnh để trả lời câu hỏi này 

\(x^4-5x^2y^2+4y^4\)

\(=\left(x^2\right)^2-2x^22y^2+\left(2y^2\right)^2-x^2y^2\)

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

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