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

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

a. \(x^3-3x^2+5x-6:\left(x^2-x+3\right)=x-2\)

b. \(8x^3-1:2x-1=4x^2+2x+1\)

ĐKXĐ: \(\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}>0\Leftrightarrow x-7>0\Leftrightarrow x>7\) (vì \(2\sqrt{15}-\sqrt{59}>0\)

ĐKXĐ: \(0\le x,y\le2012\)

Dễ thấy hệ pt trên là hệ pt đối xứng nên

\(x=y\)

Suy ra \(\sqrt{x}+\sqrt{2012-x}=\sqrt{2012}\Leftrightarrow2012+2\sqrt{2012x-x^2}=2012\) \(\Leftrightarrow\sqrt{2012x-x^2}=0\Leftrightarrow x\left(2012-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2012\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\\y=2012\end{matrix}\right.\)

Vậy nghiệm của hệ pt là (0;0) , (2012;2012)

a. \(9\left(x-3\right)^2=4\left(x+2\right)^2\Leftrightarrow\left(3\left(x-3\right)\right)^2-\left(2\left(x+2\right)\right)^2=0\Leftrightarrow\left(3x-9+2x+4\right)\left(3x-9-2x-4\right)=0\Leftrightarrow\left(5x-5\right)\left(x-13\right)=0\Leftrightarrow\left[{}\begin{matrix}5x-5=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=13\end{matrix}\right.\)

b. \(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\Leftrightarrow\left(x+1\right)^2=4\left(\left(x-1\right)^2\right)^2\Leftrightarrow\left(x+1\right)^2-\left(2\left(x-1\right)^2\right)^2=0\Leftrightarrow\left(x+1+2x^2-4x+2\right)\left(x+1-2x^2+4x-2\right)=0\Leftrightarrow\left(2x^2-3x+3\right)\left(-2x^2+5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x^2-3x+3=0\\-2x^2+5x-1=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5\pm\sqrt{17}}{4}\)

Vì \(2x^2-3x+3>0\)

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

e/ \(x^2+2x+1-16y^2=\left(x+1\right)^2-\left(4y\right)^2=\left(x+1-4y\right)\left(x+1+4y\right)\)

a. \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-x^2+9-6=0\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}\)

b. \(9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\Leftrightarrow9x^2-4-12x^2+23x-10=0\Leftrightarrow-3x^2+23x-14=0\Leftrightarrow\left(x-7\right)\left(-3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-7=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

c. \(x^2\left(x+3\right)-x^2-3x=0\Leftrightarrow x^3+2x^2-3x=0\Leftrightarrow x\left(x^2+2x-3\right)=0\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-3\end{matrix}\right.\)

a. \(x^2+6x-y^2+9=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)

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

c. \(2xy+3z+6y+xz=2xy+6y+3z+xz=2y\left(x+3\right)+z\left(3+x\right)=\left(2y+z\right)\left(x+3\right)\)

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

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

c. \(4x^4-37x^2+9=\left(x^2-9\right)\left(4x^2-1\right)\)

d. \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=x^4+2x^3+x^2+4x^2+4x-12=x^4+2x^3+5x^2+4x-12=\left(x-1\right)\left(x^3+3x^2+8x+12\right)=\left(x-1\right)\left(x+2\right)\left(x^2-x+6\right)\)

e. \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=x^4+20x^3+124x^2+240x+128=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)