\(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)\)

\(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2x-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)=\left(3x-2x+2\right)\left(3x+2y\right)\)

\(9x^2+6x-4y^2+4y\)

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

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

\(\left(x^2+6x-1\right)^2+2x^2+x^4+2\left(x^2+6x-1\right)\left(x^2+1\right)\)

\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+\left(x^2+1\right)^2-1=\left(x^2+6x-1+x^2+1\right)^2-1=\left(2x^2+6x\right)^2-1=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)

\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+x^4+2x^2\)

\(=\left(x^2+6x-1\right)\left(x^2+6x-1+2x^2+2\right)+x^4+2x^2\)

\(=\left(x^2+6x-1\right)\left(3x^2+6x+1\right)+x^4+2x^2\)

\(=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

Ta có: (x²+6x-5)(x²+6x+3)-20

      = [(x²+6x-1)-4][(x²+6x-1)+4]-20

      = (x²+6x-1)²-16-20

      = (x²+6x-1)²-36

      = (x²+6x-7)(x²+6x-5)

      = (x+7)(x-1)(x²+6x-5)

\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)\\ =\left(x^2+6x-1\right)^2-16-20\\ =\left(x^2+6x-1\right)^2-36\\ =\left(x^2+6x-1-6\right)\left(x^2+6x-1+6\right)\\ =\left(x^2+6x-7\right)\left(x^2+6x+5\right)\\ =\left(x-1\right)\left(x+7\right)\left(x+1\right)\left(x+5\right)\)

\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)-20\)

\(=\left(x^2+6x\right)^2-2\left(x^2+6x\right)-35\)

\(=\left(x^2+6x-7\right)\left(x^2+6x+5\right)\)

\(=\left(x+7\right)\left(x-1\right)\left(x+1\right)\left(x+5\right)\)

\(=\left(x+3\right)^6-y^6\\ =\left[\left(x+3\right)^3-y^3\right]\left[\left(x+3\right)^3+y^3\right]\\ =\left(x+3-y\right)\left[\left(x+3\right)^2+y\left(x+3\right)+y^2\right]\left(x+3+y\right)\left[\left(x+3\right)^2-y\left(x+3\right)+y^2\right]\\ =\left(x+y+3\right)\left(x-y+3\right)\left(x^2+6x+9+xy+3y+y^2\right)\left(x^2+6x+9-xy-3y+y^2\right)\)

\(\left(x^2+6x+9\right)^3-\left(y^2\right)^3=\left(x^2+6x+9-y^2\right)\left[\left(x^2+6x+9\right)^2+\left(x^2+6x+9\right)y^2+y^4\right]\)

\(=\left[\left(x+3\right)^2-y^2\right]\left\{\left[\left(x^2+6x+9\right)^2+2\left(x^2+6x+9\right)y^2+y^4\right]-\left(x^2+6x+9\right)y^2\right\}\)

\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)^2-\left(x+3\right)^2y^2\right]\)

\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)-\left(x+3\right)y\right]\left(x^2+6x+9+y^2\right)+\left(x+3\right)y\)

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

\(\left(x^2-2x-6\right)\left(x^2-2x-11\right)+6\)

\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+66+6\)

\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+72\)

\(=\left(x^2-2x-8\right)\left(x^2-2x-9\right)\)

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

\(x^2-2x-24\)

\(=x^2-6x+4x-24\)

\(=x(x-6)+4(x-6)\)

\(=(x+4)(x-6)\)

\(x^2-2x-24\\ =x^2-2x+1-25\\ =\left(x-1\right)^2-5^2\\ =\left(x-1-5\right)\left(x-1+5\right)\\ =\left(x-6\right)\left(x+4\right)\)

\(x^2-2x-24=\left(x-6\right)\left(x+4\right)\)

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

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

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

\(=(x+3)(x+5)(x-1)(x+1)\)

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

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

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

\(=\left(x^2-1\right)\left(x+3\right)\left(x+5\right)\)