9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

<=>9x²-18x+9+y²-6y+9+2z²+4z+2=0

<=>(3x-3)²+(y-3)²+2(z²+2z+1)=0

<=>(3x-3)²+(y-3)²+2(z+1)²=0

=>3x-3=0 và y-3=0 và z+1=0

<=>x=1 và y=3 và z=-1

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Suy ra hoặc \(3x-3=0\Leftrightarrow x=1\)

            hoặc \(y-3=0\Leftrightarrow y=3\)

            hoặc \(z+1=0\Leftrightarrow z=-1\)

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)(*)

Vì \(\left(x-1\right)\ge0;\left(y-3\right)^2\ge0;\left(z+1\right)^2\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\y=3\\z=-1\end{cases}}}\)

pt ⇔ ( 9x² - 18x + 9 ) + ( y² - 6y + 9 ) + ( 2z² + 4z + 2 ) = 0

    ⇔ 9( x² - 2x + 1 ) + ( y - 3 )² + 2( z² + 2z + 1 ) = 0

    ⇔ 9( x - 1 )² + ( y - 3 )² + 2( z + 1 )² = 0

Vì \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\2\left(z+1\right)^2\ge0\forall z\end{cases}}\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)

Vậy 

2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0

=> 2x=3y;  5y=2z ;  3z=5x => x/3=y/2; y/2=z/5

=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31

      x/3 = 3y/6=2z/10 = (x-3y+2z)/7

=>  (12x+5y-3z)/ (x-3y+2z)=31/7

9x^2+ y^2 + 2z^2 - 18x + 4z - 6y + 20 = 0

<=>9x²-18x+9+y²-6y+9+2z²+4z+2=0

<=>(3x-3)²+(y-3)²+2.(z²+2z+1)=0

<=>(3x-3)²+(y-3)²+2.(z+1)²=0

<=>3x-3=0 và y-3=0 và z+1=0

<=>x=1 và y=3 và z=-1