\(4x^2+64y^2\\ =\left(2x\right)^2+\left(8y\right)^2\\ =4x^2+32xy+64y^2\)

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

\(x^2-4x=0\\ =x\left(x-4\right)=0\)

\(=>x=0\)

\(x=4\)

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

\(4x^4+1 =(2x^2)^2+4x^2+1-4x^2 =(2x^2+1)^2-(2x)^2 =(2x^2+1+2x)(2x^2+1-2x)\)

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

\(x^3-6x^2y+12xy^2-8y^3=\left(x-2\right)^3\)

\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

\(x^3+64=x^3+4^3=(x+4)(x^2-3x+16)\)

\(a)2x+10y=2(x+5y) \)

\(b)x^2+xy+x=x(x+y+1) \)

\(c)3x^2y-6xy+12xy^2=3xy(x-2+4y)\)

\(d)y(x-2)-2x(x-2)=(y-2x)(x-2)\)

\(e)\)\(x^2-2x-15\\= x^2-5x+3x-15\\ =x\left(x-3\right)+5\left(x-3\right)\\ =\left(x+5\right)\left(x-3\right)\)

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

\(5x^2+10xy+5y^2\\ =5\left(x^2+2xy+y^2\right)\\ =5\left(x+y\right)^2\)